Answer:
(a) 0.48006
(b) 0.3251
(c) 0.19489
(d) 97.5% of people consumed less than 56 gallons of bottled water.
Step-by-step explanation:
We are given that the per capita consumption of bottled water is approximately normally distributed with a mean of 34.5 and a standard deviation of 11 gallons.
Let X = per capita consumption of bottled water
So, X ~ N([tex]\mu = 34.5,\sigma^{2} =11^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) Probability that someone consumed more than 35 gallons of bottled water = P(X > 35)
P(X > 35) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{35-34.5}{11}[/tex] ) = P(Z > 0.05) = 1 - P(Z [tex]\leq[/tex] 0.05)
= 1 - 0.51994 = 0.48006
(b) Probability that someone consumed between 25 and 35 gallons of bottled water = P(25 < X < 35)
P(25 < X < 35) = P(X < 35) - P(X [tex]\leq[/tex] 25)
P(X < 35) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{35-34.5}{11}[/tex] ) = P(Z < 0.05) = 0.51994
P(X [tex]\leq[/tex] 25) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{25-34.5}{11}[/tex] ) = P(Z [tex]\leq[/tex] -0.86) = 1 - P(Z < 0.86)
= 1 - 0.80511 = 0.19489
Therefore, P(25 < X < 35) = 0.51994 - 0.19489 = 0.3251
(c) Probability that someone consumed less than 25 gallons of bottled water = P(X < 25)
P(X < 25) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{25-34.5}{11}[/tex] ) = P(Z < -0.86) = 1 - P(Z [tex]\leq[/tex] 0.86)
= 1 - 0.80511 = 0.19489
(d) We have to find that 97.5% of people consumed less than how many gallons of bottled water, which means ;
P(X < x) = 0.975
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-34.5}{11}[/tex] ) = 0.975
P(Z > [tex]\frac{x-34.5}{11}[/tex] ) = 0.975
Now in the z table the critical value of X which have an area less than 0.975 is 1.96, i.e.;
[tex]\frac{x-34.5}{11}[/tex] = 1.96
[tex]x[/tex] - 34.5 = [tex]1.96 \times 11[/tex]
[tex]x[/tex] = 34.5 + 21.56 = 56.06 ≈ 56 gallons of bottled water
So, 97.5% of people consumed less than 56 gallons of bottled water.
a. The probability that someone consumed more than 35 gallons of bottled water is approximately 52.12%. b. The probability that someone consumed between 25 and 35 gallons of bottled water is approximately 9.58%. c. The probability that someone consumed less than 25 gallons of bottled water is approximately 42.51%. d. 97.5% of people consumed less than approximately 57.56 gallons of bottled water.
Explanation:a. To find the probability that someone consumed more than 35 gallons of bottled water, we need to standardize the value of 35 using the formula z = (x - mean) / standard deviation, where x is the value we want to find the probability for. In this case, x = 35. Plugging in the values, we get z = (35 - 34.5) / 11 = 0.045. Using a standard normal distribution table, the probability (area under the curve) to the right of 0.045 is approximately 0.5212. Therefore, the probability that someone consumed more than 35 gallons of bottled water is 0.5212 or 52.12%.
b. To find the probability that someone consumed between 25 and 35 gallons of bottled water, we need to find the probability to the right of 25 (P(x > 25)) and subtract the probability to the right of 35 (P(x > 35)). Following the same process as in part (a), we find that P(x > 25) ≈ 0.5746 and P(x > 35) ≈ 0.4788. Therefore, the probability that someone consumed between 25 and 35 gallons of bottled water is approximately 0.5746 - 0.4788 = 0.0958 or 9.58%.
c. To find the probability that someone consumed less than 25 gallons of bottled water, we need to find the probability to the left of 25 (P(x < 25)). Using the standard normal distribution table, we find that P(x < 25) ≈ 0.4251. Therefore, the probability that someone consumed less than 25 gallons of bottled water is approximately 0.4251 or 42.51%.
d. To find the value at which 97.5% of people consumed less than, we need to find the z-score that corresponds to the 97.5th percentile of the standard normal distribution. Using the standard normal distribution table, we find that the z-score corresponding to the 97.5th percentile is approximately 1.96. Now we can use the formula z = (x - mean) / standard deviation to solve for x. Plugging in the values, we get 1.96 = (x - 34.5) / 11. Solving for x, we find x ≈ 1.96 * 11 + 34.5 ≈ 57.56. Therefore, 97.5% of people consumed less than approximately 57.56 gallons of bottled water.
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The research question below describes a relationship between two quantitative variables. Which variable should be plotted on the horizontal X-axis? Is the sales price of a townhouse in San Francisco related to the number of square feet in the townhouse?
Answer:
number of square feet in the townhouse.
Step-by-step explanation:
When the relationship of two quantitative variables is assessed through plot then the scatter plot is made. Scatter plot shows relationship between two qualitative variables by showing dependent variable on y axis and independent variable on x axis.
When the relation between sale price of house and number of square feet in house is assessed then the sale price is dependent variable and square feet in house. Sale price is dependent variable because the price of house depends on the number of square feet of house. Thus, number of square feet in the town house will be plotted on the horizontal X-axis.
The director of a MPA program finds the incoming 200 students in the department with a mean GPA of 3.04 and a standard deviation of .57. Assuming that students' GAPs are normally distributed, what is the range of GPA of 95% students
Answer:
The range of GPA of 95% students is from 1.9 to 3.18.
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 3.04
Standard deviation = 0.57
Assuming that students' GAPs are normally distributed, what is the range of GPA of 95% students
Within 2 standard deviations of the mean
3.04 - 2*0.57 = 1.9
3.04 + 2*0.57 = 3.18
The range of GPA of 95% students is from 1.9 to 3.18.
Some parts of California are particularly earthquake-prone. Suppose that in one metropolitan area, 31% of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random. Let X denote the number among the four who have earthquake insurance.
(a) Find the probability distribution of X. [Hint: Let S denote a homeowner that has insurance and F one who does not. Then one possible outcome is SFSS, with probability (0.31)(0.69)(0.31)(0.31) and associated X value 3. There are 15 other outcomes.] (Round your answers to four decimal places.)
(b) What is the most likely value for X?
(c) What is the probability that at least two of the four selected have earthquake insurance? (Round your answer to four decimal places.)
Answer:
a.) 0.0822
b.) 1
c.) 0.3659
Step-by-step explanation:
Probability distribution formula is often denoted by :
P(X=r) = nCr × p^r × q^n-r
Where n = total number of samples
r = number of successful outcome of sample
p = probability of success
q = probability of failure.
If we take 4 samples,
3 of this 4 samples are successful
the success rate =S= 31% = 0.31
Failure rate = F= 0.69
a.) Then probability distribution of X becomes:
P(X=3) = 4C3 × 0.31³ × 0.69¹
P(X=3) = 0.0822 (4d.p),
b. Most likely value of X = expected value = np
= 4 × 0.31
= 1.24 ≈ 1
c.) probability that at least 2 out of the 4 have insurance = Probability that 2 have insurance) + probability that 3 have insurance + probability that 4 have insurance.
P(X=2) = 4C2 × 0.31² × 0.69² = 0.2745
P(X=3), as calculated earlier = 0.0822
P(X=4) = 4C4 × 0.31^4 × 0.69^0 = 0.0092
Total probability of having at least 2 out of those 4 insured = 0.2745 + 0.0822 + 0.0092 =0.3659.
A market research firm conducts telephone surveys with a 44% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions?
Answer:
So, the probability is P=0.9953.
Step-by-step explanation:
We know that a market research firm conducts telephone surveys with a 44% historical response rate.
We get that:
[tex]p=44\%=0.44=\mu_{\hat{x}}\\\\n=400\\\\\hat{p}=\frac{150}{400}=0.375\\\\[/tex]
We calculate the standar deviation:
[tex]\sigma_{\hat{p}}=\sqrt{\frac{0.44(1-0.44)}{400}}\\\\\sigma_{\hat{p}}=0.025[/tex]
So, we get
[tex]z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma{\hat{p}}}\\\\z=\frac{0.375-0.44}{0.025}}\\\\z=-2.6[/tex]
We use a probability table to calculate it
[tex]P(\hat{p}>0.375)=P(z>-2.6)=1-P(z<-2.6)=1-0.0047=0.9953[/tex]
So, the probability is P=0.9953.
To find the probability of at least 150 of 400 individuals responding given a 44% response rate, calculate the mean and standard deviation and then find the z-score for 150 responses. The probability of getting at least 150 responses is the area to the right of the z-score in the standard normal distribution.
Explanation:The question asks for the probability that in a new sample of 400 telephone numbers, at least 150 individuals will respond, given a historical response rate of 44%. To determine this probability, we can approximate the binomial distribution to a normal distribution because the sample size is large (n=400).
First, we calculate the mean (μ) and the standard deviation (σ) for the number of responses. The mean is given by μ = np = 400 × 0.44 = 176. The standard deviation is σ = √(np(1-p)) = √(400 × 0.44 × 0.56) ≈ 9.92.
To calculate the probability of getting at least 150 responses, we would find the z-score for 150, which is z = (X - μ)/σ = (150 - 176)/9.92 ≈ -2.62. We then look up this z-score in a standard normal distribution table or use a calculator to determine the area to the right of this z-score, which represents the probability of getting more than 150 responses.
The question is related to market research and involves using statistical methods to calculate probabilities, which requires an understanding of binomial distributions and normal approximations.
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Suppose you work for Fender Guitar Company and you are responsible for testing the integrity of a new formulation of guitar strings. To perform your analysis, you randomly select 52 'high E' strings and put them into a machine that simulates string plucking thousands of times per minute. You record the number of plucks each string takes before failure and compile a dataset. You find that the average number of plucks is 5,314.4 with a standard deviation of 116.68. A 90% confidence interval for the average number of plucks to failure is (5,287.3, 5,341.5).
From the option listed below, what is the appropriate interpretation of this interval?
1) We are 90% confident that the average number of plucks to failure for all 'high E' strings tested is between 5,287.3 and 5,341.5
2) We cannot determine the proper interpretation of this interval.
3) We are 90% confident that the proportion of all 'high E' guitar strings fail with a rate between 5,287.3 and 5,341.5
4) We are certain that 90% of the average number of plucks to failure for all 'high E' strings will be between 5,287.3 and 5,341.5 5
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5
Answer:
[tex]5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30[/tex]
[tex]5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5[/tex]
So on this case the 90% confidence interval would be given by (5287.3;5341.5)
And the best intrpretation is:
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X=5314.4[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=116.68[/tex] represent the sample standard deviation
n=52 represent the sample size
90% confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex] df = n-1= 52-1=51[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,51)".And we see that [tex]t_{\alpha/2}=1.675[/tex]
Now we have everything in order to replace into formula (1):
[tex]5314.4-1.675\frac{116.68}{\sqrt{52}}=5287.30[/tex]
[tex]5314.4+1.675\frac{116.68}{\sqrt{52}}=5341.5[/tex]
So on this case the 90% confidence interval would be given by (5287.3;5341.5)
And the best intrpretation is:
5) We are 90% confident that the average number of plucks to failure for all 'high E' strings is between 5,287.3 and 5,341.5
Suppose that we have collected a sample of 8 observations with values 12.6, 12.9, 13.4, 12.3, 13.6, 13.5, 12.6, and 13.1. What are the observed sample mean, observed sample variance, and observed sample standard deviation
Answer:
The sample mean is 13.
The sample variance is 0.2286.
The sample standard deviation is 0.4781.
Step-by-step explanation:
The sample is:
S = {12.6, 12.9, 13.4, 12.3, 13.6, 13.5, 12.6, 13.1}
The sample is of size n = 8.
The formula to compute the sample mean, sample variance and sample standard deviation are:
[tex]\bar x=\frac{1}{n} \sum x[/tex]
[tex]s^{2}=\frac{1}{n}\sum (x-\bar x)^{2} \\s=\sqrt{\frac{1}{n}\sum (x-\bar x)^{2} }\\[/tex]
Compute the sample mean as follows:
[tex]\bar x=\frac{1}{n} \sum x\\=\frac{1}{8}(12.6+ 12.9+ 13.4+ 12.3+ 13.6+ 13.5+ 12.6+ 13.1)\\=\frac{104}{8}\\ =13[/tex]
The sample mean is 13.
Compute the sample variance as follows:
[tex]s^{2}=\frac{1}{n-1}\sum (x-\bar x)^{2} \\=\frac{1}{8-1}[(12.6-13)^{2}+(12.9-13)^{2}+(13.4-13)^{2}+...+(13.1-13)^{2}] \\=\frac{1}{7}\times1.6\\=0.2286[/tex]
The sample variance is 0.2286.
Compute the sample standard deviation as follows:
[tex]s=\sqrt{s^{2}}\\=\sqrt{0.2286}\\=0.4781[/tex]
The sample standard deviation is 0.4781.
Answer:
The sample mean is 13.
The sample variance is 0.2286.
The sample standard deviation is 0.4781.
Step-by-step explanation:
Good luck
Starting from rest, a DVD steadily accelerates to 500 rpm in 1.0 s, rotates at this angular speed for 3.0 s, then steadily decelerates to a halt in 2.0 s. How many revolutions does it make
Answer:
37.5 revolutions
Step-by-step explanation:
The average rotation speed for the first second and for the last two seconds is:
[tex]V_1 = \frac{0+500}{2}\\ V_1 = 250\ rpm[/tex]
For the next 3.0 seconds, the rotation speed is V = 500 rpm.
The total number of revolutions, converting rpm to rps, is given by:
[tex]n=\frac{1*V_1+3*V+2*V_1}{60}\\n=\frac{1*250+3*500*2*250}{60}\\n=37.5\ revolutions[/tex]
The DVD makes 37.5 revolutions.
If, based on a sample size of 950, a political candidate finds that 563 people would vote for him in a two-person race, what is the 99% confidence interval for his expected proportion of the vote? Would he be confident of winning based on this poll?
Answer:
The 99% confidence interval for his expected proportion of the vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 950, p = \frac{563}{950} = 0.5926[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 - 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.5516[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 + 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.6336[/tex]
The 99% confidence interval for his expected proportion of the vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.
A VCR manufacturer receives 70% of his parts from factory F1 and the rest from factory F2. Suppose that 3% of the output from F1 are defective while only 2% of the output from F2 are defective. a. What is the probability that a received part is defective? b. If a randomly chosen part is defective, what is the probability it came from factory F1?
Answer:
See the explanation.Step-by-step explanation:
Lets take 1000 output in total.
From these 1000 outputs, 700 are from F1 and 300 are from F2.
Defective from F1 is [tex]\frac{700\times3}{100} = 21[/tex] and defective from F2 is [tex]\frac{300\times2}{100} = 6[/tex].
a.
Total received part is 1000.
Total defectives are (21+6) = 27.
The probability of a received part being defective is [tex]\frac{27}{1000}[/tex].
b.
The probability of the randomly chosen defective part from F1 is [tex]\frac{21}{27} = \frac{7}{9}[/tex].
The probability that a received part is defective is 2.7%, while the probability that a randomly chosen defective part came from factory F1 is 77.78%.
Explanation:To find the probability that a received part is defective, we need to consider the probabilities of receiving a defective part from each factory. Let's assume that the VCR manufacturer receives 100 parts. From factory F1, 70% of the parts are received, which means 70 parts. The probability of a part from F1 being defective is 3%, so the number of defective parts from F1 is 70 * (3/100) = 2.1. From factory F2, which accounts for the remaining 30% of the parts (30 parts), the probability of a part being defective is 2%, resulting in 30 * (2/100) = 0.6 defective parts. Therefore, the total number of defective parts is 2.1 + 0.6 = 2.7. Since there are 100 parts in total, the probability that a received part is defective is 2.7/100 = 0.027, or 2.7%.
To find the probability that a randomly chosen defective part came from factory F1, we can use the concept of conditional probability. The probability of a part coming from F1 given that it is defective can be found using the formula P(F1|Defective) = P(F1 and Defective) / P(Defective). We already know that P(F1 and Defective) = 2.1/100 and P(Defective) = 2.7/100. Substituting these values, we get P(F1|Defective) = (2.1/100) / (2.7/100) = 2.1/2.7 = 0.7778, or 77.78%.
You collect 100 samples from a large butterfly population. Fifty specimens are dark brown, 20 are speckled, and 30 are white. Coloration in this species of butterfly is controlled by one gene locus: BB individuals are brown, Bb are speckled, and bb are white. What are the allele frequencies for the coloration gene in this population
Answer:
Total Allele: 100
BB = 100 ; Bb = 40 ; bb = 60
[tex]\rho = 0.6\\\delta = 0.4[/tex]
Step-by-step explanation:
There are two allele in each gene. Since we have 100 samples of butterfly genes, the total number of allele are 100 x 2 = 200.
For each species:
Dark Brown (BB) → 50 x 2 = 100 allele
Speckled (Bb) → 20 x 2 = 40 allele
White (bb) → 30 x 2 = 60 allele
So, if we let [tex]\rho[/tex] be the frequency of the B allele and [tex]\delta[/tex] be the frequency of the b alleles, then:
[tex]\rho = \frac{ ((50 \times 2) + 20)}{200} \\\ \\\rho = \frac{100+20}{200} = \frac{120}{200}\\\\\\rho = 0.6[/tex]
[tex]\delta = \frac{((30 x 2) + 20)}{200}\\\\\delta = \frac{60+20}{200} = \frac{80}{200}\\\\\delta = 0.4[/tex]
We want to get the allele frequencies for the coloration gene in the population of butterflies, we will get:
The frequency for BB (brown) is 50%The frequency for Bb (speckled) is 20%The frequency for bb (white) is 30%How to get the frequencies?
Assuming that the sample is a good representation of the butterfly population, the frequencies are just given by the quotient between the number of each type of butterflies and the total number of butterflies in the sample, times 100%.
There are 100 butterflies, 50 are dark brown, 20 are speckled, and 30 are white.
The frequency for BB (brown) is:
[tex]F_{BB} = (50/100)*100\% = 50\% [/tex]
The frequency for Bb (speckled) is:
[tex]F_{Bb} = (20/100)*100\% = 20\%[/tex]
The frequency for bb (white) is:
[tex]F_{bb} = (30/100)*100\% = 30\%[/tex]
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Derek walks along a road which can be modeled by the equation y =2x, where (0,0) represents his starting point. When he reaches the point (7, 14), he turns right, so that he is traveling perpendicular to the original road, until he stops at a point which is due east of his starting point (in other words, on the x-axis). What is the point where Derek stops? Select the correct answer below: (39,0) (38, 0) (31,0) (29, 0) (35, 0) (30,0)
Answer:
(35,0)
Step-by-step explanation:
Consider the diagram below, the starting point is given as A and the finish point given as C.
Using similar right-angle triangle, we have that:
[tex]\frac{|AM|}{|BM|}= \frac{|BM|}{|MC|}\\\frac{7}{14}= \frac{14}{x}\\7x=14 X 14\\x=196/7=28[/tex]
Therefore to find the point where Derek stops at C, we first determine the distance |AC|
|AC|=7+28=35
The Coordinates at C where Derek stops is (35,0)
Derek walks along a road described by the equation y = 2x. At the point (7,14), he turns right and walks perpendicularly to his original path until he reaches the x-axis. Upon reaching the x-axis, his stopping point is at (35,0).
Explanation:When Derek reached the point (7,14), he turned right and started walking perpendicular to the original road. Given that this road is represented by the linear equation y = 2x, a perpendicular path would have a negative reciprocal slope. Therefore, the path he took after turning is represented by y = -1/2x + b. As he turned at the point (7, 14), substituting these coordinates into the equation provides b = 17.5. Since he stopped on the x-axis where y = 0, putting this into the equation gives x = 35. So, Derek stopped at (35,0).
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Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the speeds (in miles per hour) of the 20 vehicles. 19 19 22 24 25 27 28 37 35 30 37 36 39 40 43 30 31 36 33 35 How would you use the data to do this problem?
Answer:
So, the sample mean is 31.3.
So, the sample standard deviation is 6.98.
Step-by-step explanation:
We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.
We calculate the sample mean :
[tex]\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3[/tex]
So, the sample mean is 31.3.
We use the formula for a sample standard deviation:
[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}[/tex]
Now, we calculate the sum
[tex]\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\[/tex]
Therefore, we get
[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}\\\\\sigma=\sqrt{\frac{1}{19}\cdot926.2}\\\\\sigma=6.98[/tex]
So, the sample standard deviation is 6.98.
In analyzing hits by certain bombs in a war, an area was partitioned into 553 regions, each with an area of 0.95 km2. A total of 535 bombs hit the combined area of 553 regions. Assume that we want to find the probability that a randomly selected region had exactly two hits. In applying the Poisson probability distribution formula, P(x)equalsStartFraction mu Superscript x Baseline times e Superscript negative mu Over x exclamation mark EndFraction , identify the values of mu, x, and e. Also, briefly describe what each of those symbols represents. Identify the values of mu, x, and e.
Answer:
Probability of having two hits in the same region = 0.178
mu: average number of hits per region
x: number of hits
e: mathematical constant approximately equal to 2.71828.
Step-by-step explanation:
We can describe the probability of k events with the Poisson distribution, expressed as:
[tex]P(x=k)=\frac{\mu^ke^{-\mu}}{k!}[/tex]
Being μ the expected rate of events.
If 535 bombs hit 553 regions, the expected rate of bombs per region (the events for this question) is:
[tex]\mu=\frac{\#bombs}{\#regions} =\frac{535}{553}= 0.9674[/tex]
For a region to being hit by two bombs, it has a probability of:
[tex]P(x=2)=\frac{\mu^2e^{-\mu}}{2!}=\frac{0.9674^2e^{-0.9674}}{2!}=\frac{0.9359*0.38}{2}=0.178[/tex]
According to one theory of learning, the number of items, w(t), that a person can learn after t hours of instruction is given by: w(t) = 15 3 t2, 0 ≤ t ≤ 64 Find the rate of learning at the end of eight hours of instruction.
Answer:
The rate of study is 5 items per hour.
Step-by-step explanation:
Number of items a person can learn after t hours of instruction, w(t) is given by:
[tex]w(t)=15\sqrt[3]{t^{2}}[/tex]
We want to determine the rate of learning at any time t. The rate is the derivative of w(t) with respect to time.
[tex]\frac{dw(t)}{dt} =\frac{d}{dt} 15\sqrt[3]{t^{2}}[/tex]
[tex]\frac{dw(t)}{dt} =15\frac{d}{dt} {t^{2/3}}[/tex]
[tex]\frac{dw(t)}{dt} =15X\frac{2}{3} {t^{2/3-1}}[/tex]
[tex]\frac{dw(t)}{dt} =10 {t^{-\frac{1}{3} }}=\frac{10}{t^\frac{1}{3}}[/tex]
Therefore, the rate of learning at any time t
[tex]\frac{dw(t)}{dt} =\frac{10}{t^\frac{1}{3}}[/tex]
At the end of 8 hours, t=8
[tex]\frac{dw(t)}{dt} =\frac{10}{8^\frac{1}{3}}[/tex]
[tex]\frac{dw(t)}{dt} =\frac{10}{2}[/tex]=5
The rate of study is 5 items per hour.
For its first year of operations, Marcus Corporation reported pretax accounting income of $274,800. However, because of a temporary difference in the amount of $19,200 relating to depreciation, taxable income is only $255,600. The tax rate is 39%. What amount should Marcus report as its deferred income tax liability in its balance sheet at the end of that year
Answer:
Please find attached file for complete answer solution and explanation of same question
Step-by-step explanation:
Answer:
$374,484
Step-by-step explanation:
Amount payable as income tax = 0.39 X $255,600 = $99,684
The amount Marcus receives if he deferred income tax that year = $(274,800 + 99,684) = $374,484
Before the distribution of certain statistical software, every fourth compact disk (CD) is testedfor accuracy. The testing process consists of running four independent programs and checking the results. The failure rates for the four testing programs are, respectively, 0.01, 0.03, 0.02, and 0.01.a.(4pts) What is the probability that a CD was tested and failed any test
Answer:
P(T∩E) = 0.017
Step-by-step explanation:
Since every fourth CD is tested. Thus if T is the event that represents 4 disks being tested,
P(T) = 1/4 = 0.25
Let Fi represent event of failure rate. So from the question,
P(F1) = 0.01 ; P(F2) = 0.03 ; P(F3) =0.02 ; P(F4) = 0.01
Also Let F'i represent event of success rate. And we have;
P(F'1) = 1 - 0.01 = 0.99 ; P(F'2) = 1 - 0.03 = 0.97; P(F'3) = 1 - 0.02 = 0.98; P(F'4) = 1 - 0.01 = 0.99
Since all programs run independently, the probability that all programs will run successfully is;
P(All programs to run successfully) =
P(F'1) x P(F'2) x P(F'3) x P(F'4) =
0.99 x 0.97 x 0.98 x 0.97 = 0.932
Now, that all 4 programs failed will be = 1 - 0.932 = 0.068
Let E be denote that the CD fails the test. Thus P(E) = 0.068
Now, since testing and CD's defection are independent events, the probability that one CD was tested and failed will be =P(T∩E) = P(T) x P(E)= 0.25 x 0.068 = 0.017
Final answer:
The probability that a CD fails any of the four independent tests, given individual failure rates of 0.01, 0.03, 0.02, and 0.01, is approximately 6.88%.
Explanation:
To calculate the probability that a CD fails any test, we should first understand that the probability of failing any particular test is the same as 1 minus the probability of passing that test. Given the failure rates of 0.01, 0.03, 0.02, and 0.01 for the four independent tests, the probabilities of a CD passing each test are 0.99, 0.97, 0.98, and 0.99, respectively.
The probability that the CD passes all four tests is the product of the individual probabilities of passing each test (since the tests are independent):
P(pass all tests) = 0.99 * 0.97 * 0.98 * 0.99Subtracting this from 1 gives the probability that a CD fails at least one test:
P(fail any test) = 1 - P(pass all tests)Let's perform the calculation:
P(pass all tests) = 0.99 * 0.97 * 0.98 * 0.99 ≈ 0.9312
P(fail any test) = 1 - 0.9312 ≈ 0.0688
Therefore, the probability that a CD fails any test is approximately 0.0688 or 6.88%.
Of the 50 states, 39 are currently under court order to alleviate overcrowding and poor conditions in one or more of their prisons. If a state is selected at random, find the probability that it is not currently under such a court order. Give your answer as a reduced fraction.
Answer:
11/50
Step-by-step explanation:
39 of 50 states are under court order, so 11 of 50 states are not under court order.
Final answer:
The probability that a randomly selected state is not under a court order to alleviate overcrowding and poor conditions in its prisons is 11/50.
Explanation:
To find the probability that a randomly selected state is not currently under court order to alleviate overcrowding and poor conditions in one or more of their prisons, we can use the complement rule in probability. This rule states that the probability of an event not happening is equal to one minus the probability of the event happening. Given that 39 states are under such a court order, the probability that a state is under court order is 39/50. Therefore, the probability that a state is not under court order is 1 - (39/50).
Calculating this, we get:
P(Not under court order) = 1 - (39/50) = (50/50) - (39/50) = 11/50
The probability that a randomly selected state is not currently under a court order to alleviate overcrowding and poor conditions in its prisons is 11/50, when expressed as a reduced fraction.
Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield Upper X overbarequals180.975 and Sequals143.042. Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments.
Answer:
[tex]180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370[/tex]
[tex]180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580[/tex]
So on this case the 95% confidence interval would be given by (61.370;300.580)
Step-by-step explanation:
Previous concepts
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
Solution to the problem
[tex]\bar X=180.975[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=143.042[/tex] represent the sample standard deviation
n=8 represent the sample size
The confidence interval on this case is given by:
[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}} [/tex] (1)
We can find the degrees of freedom and we got:
[tex] df = n-1= 8-1=7[/tex]
The next step would be find the value of [tex]\t_{\alpha/2}[/tex], [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex]
Using the t table with df =7, excel or a calculator we see that:
[tex]t_{\alpha/2}=2.365[/tex]
Since we have all the values we can replace:
[tex]180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370[/tex]
[tex]180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580[/tex]
So on this case the 95% confidence interval would be given by (61.370;300.580)
A penny is tossed. We observe whether it lands heads up or tails up. Suppose the penny is a fair coin; that is, the probability of heads is one-half and the probability of tails is one-half. What does this mean?A) if I flip the coin many, many times, the proportion of heads will be approximately 1/2, and this proportion will tend to
get closer and closer to 1/2 as the number of tosses increases.
B) regardless of the number of flips, half will be heads and half tails.
C) every occurrence of a head must be balanced by a tail in one of the next two or three tosses.
D) all of the above.
Answer:
So we conclude that the answer is under (B).
Step-by-step explanation:
We know that the penny is a fair coin; that is, the probability of heads is one-half and the probability of tails is one-half.
So when we throw a coin we have an equal chance of getting either a head or a tail.
So we conclude that the answer is under (B).
B) regardless of the number of flips, half will be heads and half tails.
Which of the following meets the requirements of a stratified random sample? Multiple Choice A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six people who volunteer for the sample. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six people chosen at random, without regard to age. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include six males chosen at random, without regard to age. A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.
Answer:
Correct option is D.
Step-by-step explanation:
Random sampling implies the selection of of values or individuals in a random pattern.
A stratified random sampling is a sampling method where:
First divide the entire population into homogeneous subgroups, known as strata.Then take a random sample form each of the strata such that the sample size is proportional to the size of the strata.In this case the it is provided that the population consists of 30 members, 10 under the age of 25 and 20 over the age of 25.
So the stratas are:
10 members under the age of 25.20 members over the age of 25.Now to a random sample of size 2 is selected from strata 1 and a random sample of size 4 is selected from strata 2.
This forms a stratified random sample.
Correct option is:
A population contains 10 members under the age of 25 and 20 members over the age of 25. The sample will include two people chosen at random under the age of 25 and four people chosen at random over 25.
Final answer:
The correct choice for a stratified random sample is selecting two people randomly under the age of 25 and four people randomly over 25, as it ensures that the sample is proportional and representative of the two age groups within the population.
Explanation:
To determine which option meets the requirements of a stratified random sample, let's first define it. A stratified random sample divides the population into separate groups, known as strata, and a random sample is taken from each group. The key point here is that the sample from each stratum is proportional to the size of the stratum within the whole population, ensuring the sample is representative.
Given the multiple-choice options provided:
Sampling based on volunteers is a convenience sample, not stratified.Choosing people at random without regard to age is simple random sampling.Selecting only males is not stratified, as it does not account for another relevant characteristic, such as age.Selecting two people randomly under the age of 25 and four people randomly over 25 is the correct choice, as it ensures the sample is representative of the two age groups within the population.This type of sampling accounts for the different proportions of the sub-groups (under and over 25) in the population, which is a key aspect of stratified sampling.
Suppose a computer engineer is interested in determining the average weight of a motherboard manufactured by a certain company. A summary of a large sample provided to the engineer suggest a mean weight of 11.8 ounces and an estimated standard deviation, sigma = 0.75. How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4? How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?
Answer:
We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.
We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,
Step-by-step explanation:
How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]
Now, find the margin of error(width) as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
For this item, we have:
[tex]M = 0.4, \sigma = 0.75[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 2.575*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.4\sqrt{n} = 1.93125[/tex]
[tex]\sqrt{n} = \frac{1.93125}{0.4}[/tex]
[tex]\sqrt{n} = 4.828125[/tex]
[tex]\sqrt{n}^{2} = (4.828125)^{2}[/tex]
[tex]n = 23[/tex]
We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.
How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find the margin of error(width) as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
For this item, we have:
[tex]M = 0.5, \sigma = 0.75[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.5 = 1.96*\frac{0.75}{\sqrt{n}}[/tex]
[tex]0.5\sqrt{n} = 1.47[/tex]
[tex]\sqrt{n} = \frac{1.47}{0.5}[/tex]
[tex]\sqrt{n} = 2.94[/tex]
[tex]\sqrt{n}^{2} = (2.94)^{2}[/tex]
[tex]n \cong 9[/tex]
We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,
A climber on Mount Everest is meters from the start of his trail and at elevation meters above sea level. At meters from the start, the elevation of the trail is meters above sea level. If for near , what is the approximate elevation another 8 meters along the trail
We are not given the value of
Mount Everest
Elevation
h'(x)
So we are going to based our parameters needed to solve this question o assumptions. The main thing is to understand the process in solving this question.
So here is it!.
A climber on Mount Everest is 8000 meters from the start of his trail and at elevation 10000 meters above sea level. At (x) meters from the start, the elevation of the trail is h(x) meters above sea level. If h' (x) = 0.5 for near , what is the approximate elevation another 8 meters along the trail
Answer:
10004 meters
Step-by-step explanation:
The rate of change of elevation, h' (x) = 0.5 near 8000 meters.
Thus; we can say that the elevation increases by 0.5 for each meter traveled at a given distance(x)
So, we need to determine the new elevation after 8 meters traveled from ( 8000 to 8008).Then the elevation change can now be written as:
= (0.5 × 8)
= 4 meters.
And also the new elevation will be:
10000 + 4 meters
= 10004 meters
A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall above 10.983 ounces.
Answer:
[tex]P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)[/tex]
And we can find this probability using the complement rule and with excel or the normal standard table:
[tex]P(z>1.61)=1-P(z<1.61)=1-0.946=0.054[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(10.5,0.3)[/tex]
Where [tex]\mu=10.5[/tex] and [tex]\sigma=0.3[/tex]
We are interested on this probability
[tex]P(X>10.983)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)[/tex]
And we can find this probability using the complement rule and with excel or the normal standard table:
[tex]P(z>1.61)=1-P(z<1.61)=1-0.946=0.054[/tex]
A Chinese restaurant offers 10 different lunch specials. Each weekday for one week, Fiona goes to the restaurant and selects a lunch special. How many different ways are there for her to select her lunches for the week
Answer:
10⁵
Step-by-step explanation:
10×10×10×10×10
= 10⁵ or 100,000
A survey is conducted of students enrolled during both the Winter 2013 and Winter 2014 semesters. 12% of students got the flu during Winter 2013. 18% of students came got the flu during Winter 2014. 5% of students got the flu during both years. What is the probability that a randomly selected survey participant didnt?
Answer:
The probability that a randomly selected survey participant didn't got the flew = 0.75
Step-by-step explanation:
Students got the flu during Winter 2013 = 12 %
Students got the flu during Winter 2014 = 18 %
Students got the flu during both years = 5 %
Students got the flu at least in one year = 12 + 18 - 5 = 25 %
Students didn't got the flew = 100 - 25 = 75 %
⇒ The probability that a randomly selected survey participant didn't got the flew = [tex]\frac{75}{100}[/tex]
⇒ 0.75
This is the probability that a randomly selected survey participant didn't got the flew.
Full-time college students report spending a mean of 29 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 5 hours. Complete parts (a) through (d) below.
a. If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 28 hours per week? ___(Round to four decimal places as needed.)
b. If you select a random sample of 25 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week? ___ (Round to two decimal places as needed.)
c. What assumption must you make in order to solve (a) and (b)? (choose between A through D)
A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.
B. The sample is symmetrically distributed, such that the Central Limit Theorem will likely hold.
C. The population is uniformly distributed.
D. The population is normally distributed.
d. If you select a random sample of 64 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week? ___(Round to two decimal places as needed.)
Answer:
a) 0.8413
b) 30
c) A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.
d) 29.63
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 29, \sigma = 5[/tex]
a. If you select a random sample of 25 full-time college students, what is the probability that the mean time spent on academic activities is at least 28 hours per week?
25 students, so [tex]n = 25, s = \frac{5}{\sqrt{25}} = 1[/tex]
This is 1 subtracted by the pvalue of Z when X = 28. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{28 - 29}{1}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
1 - 0.1587 = 0.8413
0.8413 is the answer.
b. If you select a random sample of 25 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week?
Value of X when Z has a pvalue of 0.84. So X when Z = 1.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1 = \frac{X - 29}{1}[/tex]
[tex]X - 29 = 1[/tex]
[tex]X = 30[/tex]
c. What assumption must you make in order to solve (a) and (b)? (choose between A through D)
Central limit theorem works if the population is normally distributed, or if the sample means are of size at least 30
So the correct answer is:
A. The population is symmetrically distributed, such that the Central Limit Theorem will likely hold for samples of size 25.
d. If you select a random sample of 64 full-time college students, there is an 84 % chance that the sample mean is less than how many hours per week?
Now we have n = 64, so [tex]s = \frac{5}{\sqrt{64}} = 0.63[/tex]
Value of X when Z has a pvalue of 0.84. So X when Z = 1.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]0.63 = \frac{X - 29}{1}[/tex]
[tex]X - 29 = 0.63[/tex]
[tex]X = 29.63[/tex]
In this statistics problem related to full-time college students' time spent on academic activities, Z scores and the Central Limit Theorem are used to find probabilities and mean values for different sample sizes. Assumptions about the symmetry of population distribution are also discussed.
Explanation:This question is about understanding and applying concepts of probability, sample means, and Central Limit Theorem in statistics.
a. The probability that the mean time is at least 28 hours can be found by calculating a Z score. The formula for Z score is (X - μ) ⸫ (σ ⸫ √n), where X is the value we are testing (28 hours in this case), μ is the population mean (29 hours), σ is the standard deviation (5 hours), and n is the sample size (25 students). After calculating the Z score, use a standard normal distribution table to find the probability.
b. To solve this, we will again use the Z score formula but in a different way. Use the given % chance and look up the corresponding Z score on a standard normal distribution table. Then use this Z score in the Z score formula to find the X value, which is the number of hours.
c. The assumption we have to make here is option A. For the Central Limit Theorem to hold, the population distribution does not have to be normal but it should be symmetric. Moreover, for sample sizes of 30 or greater, Central Limit Theorem holds regardless of the shape of the population distribution.
d. This is similar to part b but with a larger sample size (64 students). Use the same procedures to find the answer.
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A meticulous gardener is interested in the length of blades of grass on his lawn. He believes that blade length X follows a normal distribution centered on 10 mm with a variance of 2 mm.
i. Find the probability that a blade of grass is between 9.5 and 11 mm long.ii. What are the standardized values of 9.5 and 11 in the context of this distribution? Using the standardized values, confirm that you can obtain the same probability you found in (i) with the standard normal density.iii. Below which value are the shortest 2.5 percent of blade lengths found?iv. Standardize your answer from (iii).
Answer:
i) [tex] P(9.5 < X<11)[/tex]
And we can solve this problem using the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]
And we can find this probability with this difference:
[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]
ii) The z scores for this case are:
[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]
[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]
And we can check the answer with the following excel code:
=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)
iii) [tex]P(X>a)=0.975[/tex] (a)
[tex]P(X<a)=0.025[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.96<\frac{a-10}{1.414}[/tex]
And if we solve for a we got
[tex]a=10 -1.96*1.414=7.228[/tex]
So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.
iv) [tex] z = \frac{7.228-10}{1.414}= -1.96[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the blade length of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(10,\sqrt{2})[/tex]
Where [tex]\mu=10[/tex] and [tex]\sigma=1.414[/tex]
Part i
For this case we want this probability:
[tex] P(9.5 < X<11)[/tex]
And we can solve this problem using the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(9.5<X<11)=P(\frac{9.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{11-\mu}{\sigma})=P(\frac{9.5-10}{1.414}<Z<\frac{11-10}{1.414})=P(-0.354<z<0.707)[/tex]
And we can find this probability with this difference:
[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-0.354<z<0.707)=P(z<0.707)-P(z<-0.354)=0.7602-0.3617=0.3985[/tex]
Part ii
The z scores for this case are:
[tex] z_1 = \frac{9.5-10}{1.414}= -0.354[/tex]
[tex] z_2 = \frac{11-10}{1.414}= 0.707[/tex]
And we can check the answer with the following excel code:
=NORM.DIST(0.707,0,1,TRUE)-NORM.DIST(-0.354,0,1,TRUE)
Part iii
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.975[/tex] (a)
[tex]P(X<a)=0.025[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.025 of the area on the left and 0.975 of the area on the right it's z=-1.96. On this case P(Z<-1.96)=0.025 and P(z>-1.96)=0.975
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.975[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.975[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.96<\frac{a-10}{1.414}[/tex]
And if we solve for a we got
[tex]a=10 -1.96*1.414=7.228[/tex]
So the value of height that separates the bottom 2.5% of data from the top 97.5% is 7.228.
Part iv
The z score for this value is given by:
[tex] z = \frac{7.228-10}{1.414}= -1.96[/tex]
Final answer:
The question involves calculating probabilities and finding values within a normal distribution regarding the length of grass blades. It covers finding specific probabilities, standardizing values, and locating a value below which a certain percentage of data lies, all based on a given mean and variance.
Explanation:
A meticulous gardener is interested in the length of blades of grass on his lawn. He believes that blade length X follows a normal distribution centered on 10 mm with a variance of 2 mm.
Find the probability that a blade of grass is between 9.5 and 11 mm long.
What are the standardized values of 9.5 and 11 in the context of this distribution? Using the standardized values, confirm that you can obtain the same probability you found in (i) with the standard normal density.
Below which value is the shortest 2.5 percent of blade lengths found?
Standardize your answer from (iii).
The standard deviation (sqrt(variance)) is sqrt(2) mm. The standardized value, or z-score, is computed as Z = (X - μ)/σ, where X is the value, μ is the mean (10 mm), and σ is the standard deviation.
For X = 9.5, Z = (9.5 - 10) / sqrt(2) = -0.3536.
For X = 11, Z = (11 - 10) / sqrt(2) = 0.7071.
To find the probability between 9.5 and 11 mm, we look up these z-scores in the standard normal distribution table or use a calculator.
To find the value below which the shortest 2.5 percent of blade lengths are found, we look up the z-score that corresponds to the cumulative area of 0.025 in the standard normal distribution table. Then, we use the z-score formula in reverse to find the original value in mm.
A plane delivers two types of cargo between two destinations. Each crate of cargo I is 9 cubic feet in volume and 187 pounds in weight, and earns $30 in revenue. Each crate of cargo II is 9 cubic feet in volume and 374 pounds in weight, and earns $45 in revenue. The plane has available at most 540 cubic feet and 14,212 pounds for the crates. Finally, at least twice the number of crates of I as II must be shipped. Find the number of crates of each cargo to ship in order to maximize revenue. Find the maximum revenue. crates of cargo I crates of cargo II maximum revenue $
Answer:
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
Step-by-step explanation:
Given that a plane delivers two types of cargo between two destinations
Crate I Crate II
Volume 9 9
Weight 187 374
Revenue 30 45
Let X be the no of crate I and y that of crate II
Then
[tex]9x+9y\leq 540\\187x+374y\leq 14212\\x\geq 2y[/tex]
Simplify these equations to get
[tex]x+y\leq 60\\x+2y\leq 76\\x\geq 2y[/tex]
Solving we get
[tex]y\leq 16\\x\leq 46 and x\geq 32\\32\leq x\leq 46[/tex]
REvenue = 30x+45y
The feasible region would have corner points as (60,0) or (32,16) or (46,16)
Revenue for (60,0) = 1800
(32,16) = 1680
(46,16)=2100
So maximum when 46 of I grade and 16 of II grade are produced.
Max revenue = 2100
To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. The problem can be solved using linear programming techniques to find the optimal solution.
Explanation:To maximize revenue, we need to determine the number of crates of each cargo that should be shipped. Let's assume the number of crates of cargo I is x and the number of crates of cargo II is y.
Based on the given information, the constraints for the problem are:
Volume constraint: 9x + 9y ≤ 540Weight constraint: 187x + 374y ≤ 14,212Relationship constraint: x ≥ 2yTo find the maximum revenue, we need to maximize the objective function: Revenue = 30x + 45y.
The problem can be solved using linear programming techniques, such as graphical or simplex method, to find the optimal solution. However, since these methods require plotting and iterations, the detailed calculations are beyond the scope of this response. The optimal solution will provide the values of x and y, which can be used to determine the maximum revenue.
Learn more about Optimizing revenue for cargo shipment here:https://brainly.com/question/34408399
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The number of square feet per house are normally distributed with a population standard deviation of 137 square feet and an unknown population mean. A random sample of 19 houses is taken and results in a sample mean of 1350 square feet. Find the margin of error for a 80% confidence interval for the population mean. z0.10z0.10z0.05z0.05z0.025z0.025z0.01z0.01z0.005z0.005 1.2821.6451.9602.3262.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.
Answer:
The MOE for 80% confidence interval for μ is 5.59.
Step-by-step explanation:
The random variable X is defined as the number of square feet per house.
The random variable X is Normally distributed with mean μ and standard deviation σ = 137.
The margin of error for a (1 - α) % confidence interval for population mean is:
[tex]MOE=z_{\alpha /2}\times\frac{\sigma}{\sqrt{n}}[/tex]
Given:
n = 19
σ = 137
[tex]z_{\alpha /2}=z_{0.20/2}=z_{0.10}=1.282[/tex]
Compute MOE for 80% confidence interval for μ as follows:
[tex]MOE=1.282\times\frac{137}{\sqrt{19}}=1.282\times4.36=5.58952\approx5.59[/tex]
Thus, the MOE for 80% confidence interval for μ is 5.59.
Which of the following random variables is geometric? The number of 1s in a row of 50 random digits. The number of tails when a coin is tossed 60 times. The number of digits in a randomly selected row until a 1 is found. The number of diamond cards obtained in a seven-card deal-out of a shuffled deck of 52 cards. The number of 1s when rolling a die 5 times.
Answer:
The number of digits in a randomly selected row until a 1 is found.
Final answer:
The random variable describing 'The number of digits in a randomly selected row until a 1 is found' is geometric, as it fits the criteria of a geometric distribution, which includes the number of trials needed for the first success, with constant success probability and independent trials.
Explanation:
The student has asked which of the given random variables is geometric. Among the options provided, the one describing a geometric random variable is "The number of digits in a randomly selected row until a 1 is found." A geometric distribution is concerned with the number of Bernoulli trials required to get the first success. In this case, obtaining a '1' when randomly selecting digits can be considered a success, and all trials are independent with the probability of success (getting a '1') remaining constant with each trial.
Other options such as the number of 1s in a row of 50 random digits or the number of tails when a coin is tossed 60 times describe binomial random variables, where we are interested in the number of successes within a fixed number of trials, not the trial number of the first success.