Answer:
a) X1+X2 have distribution Bi(n1+n2, p)
b)
P(X1+X2 = 1 | X2 = 0) = np(1-p)ⁿ¹⁻¹
P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹
P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
Step-by-step explanation:
Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).
b)
Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).
If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹
If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹
Thus,
P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹
An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58 type Q batteries. The mean voltage is measured as 8.54 for the type K batteries with a standard deviation of 0.225, and the mean voltage is 8.69 for type Q batteries with a standard deviation of 0.725. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1 be the true mean voltage for type K batteries and μ2 be the true mean voltage for type Q batteries.Use a 0.1 level of significance.
Answer:
Hypothesis Test states that we will accept null hypothesis.
Step-by-step explanation:
We are given that an engineer is comparing voltages for two types of batteries (K and Q).
where, [tex]\mu_1[/tex] = true mean voltage for type K batteries.
[tex]\mu_2[/tex] = true mean voltage for type Q batteries.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 = \mu_2[/tex] {mean voltage for these two types of
batteries is same}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu_1 \neq \mu_2[/tex] {mean voltage for these two types of
batteries is different]
The test statistics we use here will be :
[tex]\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows [tex]t_n__1 + n_2 -2[/tex]
where, [tex]X_1bar[/tex] = 8.54 and [tex]X_2bar[/tex] = 8.69
[tex]s_1[/tex] = 0.225 and [tex]s_2[/tex] = 0.725
[tex]n_1[/tex] = 37 and [tex]n_2[/tex] = 58
[tex]s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }[/tex] = 0.585 Here, we use t test statistics because we know nothing about population standard deviations.
Test statistics = [tex]\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }[/tex] follows [tex]t_9_3[/tex]
= -1.219
At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that mean voltage for these two types of batteries is same.
The time taken to deliver a pizza has a uniform probability distribution from 20 minutes to 60 minutes. What is the probability that the time to deliver a pizza is at least 32 minutes?
The results on a certain blood test performed in a medical laboratory are known to be approximately normally distributed, with m=60 and s=18.
a. What percentage of the results are above 45?
b. What percentage of the results are below 85?
c. What percentage of the results are between 75 and 90?
d. What percentage of the results are outside the "healthy range" of 20 to 100?
Answer:
(1) The probability that the time to deliver a pizza is at least 32 minutes is 0.70.
(2a) The percentage of results more than 45 is 79.67%.
(2b) The percentage of results less than 85 is 91.77%.
(2c) The percentage of results are between 75 and 90 is 15.58%.
(2d) The percentage of results outside the healthy range 20 to 100 is 2.64%.
Step-by-step explanation:
(1)
Let Y = the time taken to deliver a pizza.
The random variable Y follows a Uniform distribution, U (20, 60).
The probability distribution function of a Uniform distribution is:
[tex]f(x)=\left \{ {{\frac{1}{b-a};\ x\in [a, b] } \atop {0};\ otherwise} \right.[/tex]
Compute the probability that the time to deliver a pizza is at least 32 minutes as follows:
[tex]P(Y\geq 32)=\int\limits^{60}_{32} {\frac{1}{b-a} } \, dx \\=\frac{1}{60-20} \int\limits^{60}_{32} {1 } \, dx\\=\frac{1}{40}\times[x]^{60}_{32}\\=\frac{1}{40}\times[60-32]\\=0.70[/tex]
Thus, the probability that the time to deliver a pizza is at least 32 minutes is 0.70.
(2)
Let X = results of a certain blood test.
It is provided that the random variable X follows a Normal distribution with parameters [tex]\mu = 60[/tex] and [tex]s = 18[/tex].
The probabilities of a Normal distribution are computed by converting the raw scores to z-scores.
The z-scores follows a Standard normal distribution, N (0, 1).
(a)
Compute the probability that the results are more than 45 as follows:
[tex]P(X>45)=P(\frac{X-\mu}{\sigma}> \frac{45-60}{18})=P(Z>-0.833)=P(Z<0.833)=0.7967[/tex]
The percentage of results more than 45 is: [tex]0.7967\times100=79.67\%[/tex]
Thus, the percentage of results more than 45 is 79.67%.
(b)
Compute the probability that the results are less than 85 as follows:
[tex]P(X<85)=P(\frac{X-\mu}{\sigma}< \frac{85-60}{18})=P(Z<1.389)=0.9177[/tex]
The percentage of results less than 85 is: [tex]0.9177\times100=91.77\%[/tex]
Thus, the percentage of results less than 85 is 91.77%.
(c)
Compute the probability that the results are between 75 and 90 as follows:
[tex]P(75<X<90)=P(\frac{75-60}{18}<\frac{X-\mu}{\sigma}< \frac{90-60}{18})\\=P(0.833<Z<1.67)\\=P(Z<1.67)-P(Z<0.833)\\=0.9525-0.7967\\=0.1558[/tex]
The percentage of results are between 75 and 90 is: [tex]0.1558\times100=15.58\%[/tex]
Thus, the percentage of results are between 75 and 90 is 15.58%.
(d)
Compute the probability that the results are between 20 and 100 as follows:
[tex]P(20<X<100)=P(\frac{20-60}{18}<\frac{X-\mu}{\sigma}< \frac{100-60}{18})\\=P(-2.22<Z<2.22)\\=P(Z<2.22)-P(Z<-2.22)\\=0.9868-0.0132\\=0.9736[/tex]
Then the probability that the results outside the range 20 to 100 is: [tex]1-0.9736=0.0264[/tex].
The percentage of results outside the range 20 to 100 is: [tex]0.0264\times100=2.64\%[/tex]
Thus, the percentage of results outside the healthy range 20 to 100 is 2.64%.
A high school baseball player has a 0.212 batting average. In one game, he gets 9 at bats. What is the probability he will get at least 2 hits in the game?
Answer:
59.92% probability he will get at least 2 hits in the game.
Step-by-step explanation:
For each at bat, there are only two possible outcomes. Either he gets a hit, or he does not. The probability of a getting a hit in each at bat is independent. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]p = 0.212[/tex]
In one game, he gets 9 at bats. What is the probability he will get at least 2 hits in the game?
This is [tex]P(X \geq 2)[/tex] when [tex]n = 9[/tex]
He either gets less than two hits in the game, or he gets at least two hits. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.212)^{0}.(0.788)^{9} = 0.1171[/tex]
[tex]P(X = 1) = C_{9,1}.(0.212)^{1}.(0.788)^{8} = 0.2837[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1171 + 0.2837 = 0.4008[/tex]
Finally
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4008 = 0.5992[/tex]
59.92% probability he will get at least 2 hits in the game.
A cube-shaped water tank having 6 ft side lengths is being filled with water. The bottom is solid metal but the sides of the tank are thin glass which can only withstand a maximum force of 200 lb. How high (in ft) can the water reach before the sides shatter?
(Assume a density of water rho = 62.4 lb/ft3.
Round your answer to two decimal places.)
The distance the in which the water will reach before it shatters will be 1.03ft
Data;
Let the integral run from surface (y = 0) to maximum depth (y = d)ftLet the differential depth = dydtThe area of each depth = 6dy ft^2The Force Acting at any DepthUsing integration, the force at any depth y is 62.4y lb/ft^3
[tex]200 = \int\limits^d_0 {62.4y} \, 6dy\\ 200 = \int\limits^d_0 {374.4y} \, dy\\ 200 = \int\limits^6_0 {374.4y} \, dy\\ 200 = [374.4/2 y^2]_0^d\\200 = 187.2(d^2 - 0)\\200 = 187.2d^2\\d^2 = 200/187.2\\d = 1.03ft[/tex]
From the calculations above, the distance in which the water will reach before it shatters is 1.03ft
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(a) The mean age at death is 15 years and the standard deviation is 7 years. What percentage of the dinosaurs' ages were within 1 standard deviation of the mean? (Answer as a whole number.)
Answer:
68% of the dinosaurs' ages were within 1 standard deviation of the mean.
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.
What percentage of the dinosaurs' ages were within 1 standard deviation of the mean?
By the Empirical Rule, 68% of the dinosaurs' ages were within 1 standard deviation of the mean.
A survey regarding the communication habits of Americans was conducted. The study randomly selected 700 American citizens and asked them a series of questions about their daily communication methods. Three primary methods of communication were identified as social media (M), SMS text messaging (T), cell phone call (C). The results are summarized below.
480 people responded using M or T
300 people use all three primary forms of comunication
465 people responded using M or C
440 people responded using C or T
405 respondents use at least 2 of the primary forms of comunication
There was nobody who used both M and C but did not use T
210 respondents used only forms of communication that differed from the primary 3
385 people use both C and T
a. Create a well labeled Venn diagram representing the survey.
b. How many people in the survey used only 1 of the three primary forms of communication?
c. What is the probability that a survey respondent used SMS text messaging?
d. What is the probability that a survey respondent used exactly 2 of the 3 primary forms of communication?
e. Given that a survey respondent used at least 1 of the 3 primary forms of communication, what is the probability that they used all 3?
f. Given a respondent used social media and standard calling with a cell phone, what is the probability that they also used SMS text messaging?
Answer:
answers and explanation is shown in the attachment
Step-by-step explanation:
The detailed steps and necessary substitution is as shown in the attached files.
In a factory there are 100100 units of a certain product, 55 of which are defective. We pick three units from the 100 units at random. What is the probability that exactly one of them is defective
Answer:
There is a 33.67% probability that exactly one of them is defective.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Here, we can have different formats. For example, D-ND-ND is the same as ND-D-ND, that is, the ordering is not important. So we use the combinations formula.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Desired outcomes
One defective(one from a set of 55) and two non defective(two from a set of 45). So
[tex]D = C_{55,1}*C_{45,2} = \frac{55!}{54!1}*\frac{45!}{43!2!} = 55*45*22 = 54450[/tex]
Total outcomes
Three from a set of 100. So
[tex]T = C_{100,3} = \frac{100!}{97!3!} = 161700[/tex]
What is the probability that exactly one of them is defective
[tex]P = \frac{D}{T} = \frac{54450}{161700} = 0.3367[/tex]
There is a 33.67% probability that exactly one of them is defective.
Evaluate using long division first to write f(x) as the sum of a polynomial and a proper rational function. (Use C for the constant of integration. Remember to use absolute values where appropriate.)
Answer:
Step-by-step explanation:
Please kindly check the attached
(p(x))/(q(x))=f(x)+(r(x))/(q(x))
Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal disease, an inflammation of the gums. Only 33% of healthy people (those who have not had heart attacks) have this disease. Suppose that in a certain community heart attacks are quite rare, occurring with only 15% probability.
A. If someone has periodontal disease, what is the probability that he or she will have a heart attack?
B. If 38% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease will have a heart attack?
Final answer:
To find the probability of someone with periodontal disease having a heart attack, we can use conditional probability and Bayes' theorem. The probability of having a heart attack given that someone has periodontal disease is 79%. Using Bayes' theorem, we can calculate the probability of a person with periodontal disease having a heart attack.
Explanation:
To answer part A, we can use conditional probability. The probability of having a heart attack given that someone has periodontal disease is represented by P(heart attack | periodontal disease). According to the given information, 79% of people who have suffered a heart attack had periodontal disease. Therefore, P(heart attack | periodontal disease) = 0.79.
To answer part B, we can use Bayes' theorem. The probability of a person with periodontal disease having a heart attack is represented by P(heart attack | periodontal disease). According to the given information, the probability of having a heart attack in the community is 38%.
Therefore, P(heart attack) = 0.38. Let's use Bayes' theorem: P(heart attack | periodontal disease) = (P(periodontal disease | heart attack) × P(heart attack)) / P(periodontal disease). We know that P(periodontal disease | heart attack) = 0.79 from part A.
The probability of having periodontal disease in the community is represented by P(periodontal disease). In the given information, it says that only 33% of healthy people have periodontal disease. Therefore, P(periodontal disease) = 0.33. Substituting these values into Bayes' theorem, we can calculate P(heart attack | periodontal disease).
which point is located at (4 -2)
Answer:
A or C
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
If the odds of catching the flu among individuals who take vitamin C is 0.0342 and the odds of catching the flu among individuals not taking vitamin C is 0.2653, then the individuals not taking vitamin C are 7.7573. Times more likely to catch the flu than individuals taking vitamin C. A. True B. False
Answer:
Correct option (A).
Step-by-step explanation:
The probability of an individual catching a flu when he or she has taken vitamin C is, P (F|C) = 0.0342.
The probability of an individual catching a flu when he or she has not taken vitamin C is, P (F|C') = 0.2653.
Th ratio of individuals who caught the flu when they did not take vitamin C to those who took vitamin C is:
[tex]=\frac{P(F|C')}{P(F|C)}\\ =\frac{0.2653}{0.0342} \\=7.7573[/tex]
This implies that:
[tex]P(F|C')=7.7573\times P(F|C)[/tex]
Thus, the the individuals not taking vitamin C are 7.7573. Times more likely to catch the flu than individuals taking vitamin C.
The statement is True.
The statement that individuals not taking vitamin C are 7.7573 times higher at risk of catching flu than those taking it, is true based on the provided numbers in the question.
Explanation:To determine if the statement is true or false, we need to divide the odds of catching the flu among individuals not taking vitamin C by the odds of those taking it:
To do this, you divide 0.2653 (odds for those not taking vitamin C) by 0.0342 (odds for those taking vitamin C). This calculation results in approximately 7.7573.
Therefore, the statement 'individuals not taking vitamin C are 7.7573 times more likely to catch flu than individuals taking vitamin C is true as the calculation matches the provided number in the statement.
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Please help! Will Mark Brainliest! I'm struggling on how to get the answer so please clarify.
Question Choose the correct simplification of the expression: (3m/n)^4
Options:
81m^4/n^4
12m^4/n
3m^4/n^4
81m^4/n
Picture is posted with the problem for clarification. The subject is multiplying and dividing monomials.
Answer:
Option 1
Step-by-step explanation:
(3m/n)⁴ = (3⁴×m⁴)/n⁴
= 81m⁴/n⁴
Sophia and London work at a dry cleaners ironing shirts. Sophia can iron 40 shirts per hour, and London can iron 25 shirts per hour. Sophia and London worked a combined 11 hours and ironed 365 shirts. Write a system of equations that could be used to determine the number of hours Sophia worked and the number of hours London worked. Define the variables that you use to write the system.
We can define two variables: S represents Sophia's hours, and L represents London's hours. The first equation representing total combined work time is S + L = 11 hours. The second equation, representing total shirts completed, is 40S + 25L = 365 shirts.
Explanation:Let's define our variables as follows: S stands for the hours Sophia worked and L for the hours London worked. Since Sophia and London worked a total of 11 hours, we can write the first equation as S + L = 11. Each worker's production rate times the hours they worked will total the shirts they produced. So, our second equation is 40S + 25L = 365. These two equations can be used to find the number of hours Sophia and London worked respectively.
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Consider the bisection method to find a root for f(x) = 0 where f is a continuous function. We take the [0, 1] as the initial interval provided that f(0)f(1) < 0. We take the following practical stopping criteria: |bn − an| ≤ , = 1 × 10−8 . How many steps of the bisection method are needed to obtain an approximation to the root?
Answer: It would take approximately 27 steps to obtain an approximation of the root.
Step-by-step explanation: The bisection method is a numerical procedure to find a root to a equation continuous in an interval [a.b]. It consists in repeatedly halve the interval [a,b], keeping the half for which f(x) changes sign. The repetition will continue until a stopping criteria is reached. To find how many interactions is necessary, we can satisfy the equation:
[tex]\frac{1}{2^{n} }[/tex] · (b - a) ≤ ε, where a and b are the original interval and ε is the stopping criteria.
Substituting the parameters and using log 2 = 0,301, the number of iterations needed is approximately 27.
Identify the type of data (qualitative/quantitative) and the level of measurement for the data described below. Explain your choice. The average monthly rainfall in inches for a certain city throughout the year Are the data qualitative or quantitative?
A. Quantitative, because numerical values, found by either measuring or counting, are used to describe the data
B. Qualitative, because numerical values, found by either measuring or counting, are used to describe the data °
C. Qualitative, because descriptive terms are used to measure or classify the data 0
D. Quantitative, because descriptive terms are used to measure or classify the data
What is the data set's level of measurement?
A. Nominal, because the data are categories or labels that cannot be ranked.
B. Ordinal, because the data are categories or labels that can be ranked °
C. Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.
D. Interval, because the differences in the data can be meaningfully measured, but the data do not have a true zero point.
Answer:
Option A) Quantitative, because numerical values, found by either measuring or counting, are used to describe the data
Option C) Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.
Step-by-step explanation:
We are given the following in the question:
"The average monthly rainfall in inches for a certain city throughout the year"
Qualitative or quantitative:
Quantitative data are measures of values or counts and are expressed as numbers. Qualitative data are measures of 'types' and may be represented by a name, symbol, or a number code. They are non-parametric values.Thus, the given quantitative data is quantitative because rainfall is always measured and the output is expressed in numerical values.
Option A) Quantitative, because numerical values, found by either measuring or counting, are used to describe the data
Level of measurement:
Ordinal: These are the qualitative variable whose order plays an important role.Interval: When true zero does not exist. The negative values of such variables make sense.Ratio: When true zero exist. The negative values of such variable does not make any sense.Since,for the given data true zero exist and a negative values make no sense, thus its is ratio.
Option C) Ratio, because the differences in the data can be meaningfully measured, and the data have a true zero point.
The data described is quantitative because it is measured in numerical values. The level of measurement for this data is interval because the differences between data points can be meaningfully measured, but the data lacks a true zero point.
The data about the average monthly rainfall in inches for a certain city throughout the year is a Quantitative Data. This is because the data are numerical values which are obtained through measuring. Quantitative data is based on numbers and can be manipulated using statistical methods. The type of measurement used for this data set is Interval. This is because although the differences in the data (rainfall from one month to the next) can be meaningfully measured, the data do not have a 'true zero' point since there isn't a point where there’s absolutely no rainfall.
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The probability distribution of customers that walk into a coffee shop on any given day of the week is described by a Normal distribution with mean equal to 100 and standard deviation equal to 20. What is the probability that no more than 80 customers walk into the coffee shop next Monday
Answer:
15.87% probability that no more than 80 customers walk into the coffee shop next Monday
Step-by-step explanation:
Problems of normally distributed samples can be 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.
In this problem, we have that:
[tex]\mu = 100, \sigma = 20[/tex]
What is the probability that no more than 80 customers walk into the coffee shop next Monday?
This is the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 100}{20}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
So there is a 15.87% probability that no more than 80 customers walk into the coffee shop next Monday
A bag contains 5 red marbles, 8 blue marbles, and 3 green marbles. If a green marble is drawn you win $5, and if a red marble is drawn you win $1. Drawing a blue marble causes you to lose $1. Suppose you make 81 draws out of the bag Your chances remain the same, the number of red, blue and green marbles doesn't change from draw to draw. This corresponds to drawing how many tickets at random?
Final answer:
The number of tickets drawn at random can be calculated using the formula for the number of combinations.
Explanation:
To determine the number of tickets drawn at random in this scenario, we need to calculate the total number of possible outcomes. In this case, Maria is drawing marbles without replacement, meaning that the number of marbles in the bag decreases with each draw. We can use the formula for the number of combinations to calculate the total number of outcomes. The formula is nCr = n! / (r!(n-r)!), where n is the total number of marbles in the bag and r is the number of marbles drawn. In this case, n = 5+8+3 = 16 and r = 2. Therefore, the number of tickets drawn at random is 16C2 = 120.
Suppose that the national average for the math portion of the College Board's SAT is 540. The College Board periodically rescales the test scores such that the standard deviation is approximately 50. Answer the following questions using a bell-shaped distribution and the empirical rule for the math test scores. If required, round your answers to two decimal places. If your answer is negative use "minus sign".
(a) What percentage of students have an SAT math score greater than 615?
(b) What percentage of students have an SAT math score greater than 690?
(c) What percentage of students have an SAT math score between 465 and 540?
(d) What is the z-score for student with an SAT math score of 625?
(e) What is the z-score for a student with an SAT math score of 415?
Answer:
a) 6.68%
b) 0.135%
c) 43.32%
d) 1.7
e) -2.5
Step-by-step explanation:
a)
P(X>615)=P(z>(615-540)/50)
P(X>615)=P(z>1.5)
P(X>615)=P(0<z<∞)-P(0<z<1.5)
P(X>615)=0.5-0.4332
P(X>615)=0.0668
The percentage of students have an SAT math score greater than 615 is 6.68%
b)
P(X>690)=P(z>(690-540)/50)
P(X>690)=P(z>3)
P(X>690)=P(0<z<∞)-P(0<z<3)
P(X>690)=0.5-0.49865
P(X>690)=0.00135
The percentage of students have an SAT math score greater than 690 is 0.135%
c)
P(465<X<540)=P((465-540)/50<z<(540-540)/50))
P(465<X<540)=P(-1.5<z<0)
P(465<X<540)=0.4332
The percentage of students have an SAT math score between 465 and 540 is 43.32%
d)
z=(625-540)/50
z=85/50
z=1.7
The z-score for student with an SAT math score of 625 is 1.7.
e)
z=(415-540)/50
z=-125/50
z=-2.5
The z-score for a student with an SAT math score of 415 is -2.5.
A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, aguitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of$2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with astandard deviation of 5100. Which cost is the lowest, when compared to other instruments of the same type? Which cost isthe highest when compared to other instruments of the same type. Justify your answer.
Answer:
Lowest Cost is for Drum set, when compared to others instruments of same type.
Highest cost is for Guitar,when compared to others instruments of same type.
Order:
Guitar>Piano>Drum Set
Step-by-step explanation:
Consider the normal distribution for all cases.
Formula we are going to use is:
[tex]z=\frac{\bar x-\mu}{\sigma}[/tex]
where:
[tex]\bar x[/tex] is the purchasing cost
[tex]\mu[/tex] is the mean
[tex]\sigma[/tex] is the standard deviation
For Piano:
[tex]z=\frac{3000-4000}{2500}\\ z=-0.4[/tex]
For Guitar:
[tex]z=\frac{550-500}{200}\\ z=0.25[/tex]
For Drum Set:
[tex]z=\frac{600-700}{100}\\ z=-1[/tex]
Lowest Cost is for Drum set, when compared to others instruments of same type.
Highest cost is for Guitar,when compared to others instruments of same type.
Order:
Guitar>Piano>Drum Set
The drum set, priced at $600, costs the least compared to the average drum set price, being 1.0 standard deviation below the mean. The guitar, priced at $550, costs the most compared to the average guitar price, at 0.25 standard deviation above the mean. These conclusions are drawn using the z-score method to compare the instruments' prices to their respective averages and standard deviations.
To determine which musical instrument costs the least and the most when compared to others of the same type, we calculate how many standard deviations below or above the mean each instrument's cost falls. For the piano, guitar, and drum set, we will use the z-score formula, which is (z = (X - μ) / σ), where X is the observed value, μ is the mean, and σ is the standard deviation.
For the piano, the z-score is ($3,000 - $4,000) / $2,500 = -0.4.
For the guitar, the z-score is ($550 - $500) / $200 = 0.25.
For the drum set, the z-score is ($600 - $700) / $100 = -1.0.
The drum set's cost is the lowest compared to other drums because it is 1.0 standard deviations below the mean. The guitar's cost, being 0.25 standard deviations above the mean, is the highest compared to other guitars. Therefore, Matt's drums cost the least in comparison to his own instrument type, and Becca's guitar costs the most compared to her own instrument type.
A 20-minute consumer survey mailed to 500 adults aged 25-34 included a $5 Starbucks gift certificate. The same surveywas mailed to 500 adults aged 25-34 without the gift certificate. There were 65 responses from the first group and 45 fromthe second group. Form a 95 percent confidence interval for the difference of proportions. Does it include zero?
Answer:
[tex](0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129[/tex]
[tex](0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787[/tex]
And the 95% confidence interval would be given (0.00129;0.0787).
We are confident at 95% that the difference between the two proportions is between [tex]0.00129 \leq p_B -p_A \leq 0.0787[/tex]
And as we can see the confidence interval for the difference on this case not contains the 0.
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".
Solution to the problem
[tex]p_A[/tex] represent the real population proportion for the gift certificate
[tex]\hat p_A =\frac{65}{500}=0.13[/tex] represent the estimated proportion for the gift certificate
[tex]n_A=500[/tex] is the sample size required for the gift certificate
[tex]p_B[/tex] represent the real population proportion for without the gift certificate
[tex]\hat p_B =\frac{45}{500}=0.09[/tex] represent the estimated proportion for without the gift certificate
[tex]n_B=500[/tex] is the sample size required for Brand B
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex](0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129[/tex]
[tex](0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787[/tex]
And the 95% confidence interval would be given (0.00129;0.0787).
We are confident at 95% that the difference between the two proportions is between [tex]0.00129 \leq p_B -p_A \leq 0.0787[/tex]
And as we can see the confidence interval for the difference on this case not contains the 0.
The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight should the citation designation be established
Answer:
The weight of the citation designation should be at 4.8432 pounds.
Explanation:
Given
Mean [tex]= 3.2 pounds.[/tex]
Standard deviation[tex]= 0.8 pound.[/tex]
Step 1:
Consider 'y' as one of the top weight, that is, [tex]y = 2 \% = 2.054 pounds.[/tex]
Let 'x' be the weight of the citation designation.
[tex]y = \frac{x-mean}{standard\ deviation}[/tex]
[tex]=2.054 = \frac{x-3.2}{0.8}[/tex]
[tex]=2.054\times 0.8 = x-3.2[/tex]
[tex]=1.6432 = x-3.2[/tex]
[tex]x = 1.6432+3.2[/tex]
[tex]x = 4.8432[/tex]
Thus, at 4.8432 pounds citation designation be established.
To learn more about citation, refer:
https://brainly.com/question/18675815Final answer:
The citation catfish designation should be established at a weight of at least 4.84 pounds, which corresponds to the top 2% of a normal distribution given the mean is 3.2 pounds and the standard deviation is 0.8 pounds.
Explanation:
The weight at which a citation catfish designation should be established is calculated by identifying the z-score that corresponds to the top 2% of a normal distribution. Given that the average weight of a catfish is 3.2 pounds with a standard deviation of 0.8 pounds, we can use a z-table to find the z-score for the 98th percentile, which typically is around 2.05.
Using the z-score formula: z = (X - mean) / standard deviation, we rearrange to solve for the unknown X (catfish weight): X = z * standard deviation + mean. Plugging in the numbers: X = 2.05 * 0.8 + 3.2, we find that the citation catfish should weigh at least 4.84 pounds.
What is a real life word problem for the equation y=2x
Answer:
You want to buy some golden apples. Each golden apple cost $2. How much much does golden apples pass cost?
y = total cost
x = amount of apple
The distribution of a sample of the outside diameters of PVC pipes approximates a normal distribution. The mean is 14.0 inches, and the standard deviation is 0.1 inches. About 68% of the outside diameters lie between what two amounts? A. 13.9 and 14.1 inches B. 13.0 and 15.0 inches C. 13.8 and 14.2 inches D. 13.5 and 14.5 inches
Answer:
A. 13.9 and 14.1 inches
See explanation below.
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 outside diameters of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(14,0.1)[/tex]
Where [tex]\mu=14[/tex] and [tex]\sigma=0.1[/tex]
If we want the middle 68% of the data we need to have on the tails 16% on each one
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.84[/tex] (a)
[tex]P(X<a)=0.16[/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 0.16 of the area on the left and 0.84 of the area on the right it's z=-0.994. On this case P(Z<-0.994)=0.16 and P(z>-0.994)=0.84
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.16[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.16[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.994<\frac{a-14}{0.1}[/tex]
And if we solve for a we got
[tex]a=14 -0.994*0.1=13.9[/tex]
And since the distribution is symmetrical for the upper limit we can use z = 0.994 and we have:
[tex]z=0.994<\frac{a-14}{0.1}[/tex]
And if we solve for a we got
[tex]a=14 +0.994*0.1=14.1[/tex]
So the correct answer for this case would be:
A. 13.9 and 14.1 inches
A company that manufactures video cameras produces a basic model and a deluxe model. Over the past year, 30% of the cameras sold have been of the basic model. Of those buying the basic model, 44% purchase an extended warranty, whereas 40% of all deluxe purchasers do so. If you learn that a randomly selected purchaser has an extended warranty, how likely is it that he or she has a basic model?
Answer:
the probability is 0.32 (32%)
Step-by-step explanation:
defining the event W= has extended warranty , then
P(W)= probability of purchasing the basic model * probability of purchasing extended warranty given that has purchased the basic model + probability of purchasing the deluxe model * probability of purchasing extended warranty given that has purchased the deluxe model = 0.3 * 0.44 + 0.7 * 0.40 = 0.412
then using the theorem of Bayes for conditional probability and defining the event B= has the basic model , then
P(B/W)= P(B∩W)/P(W)= 0.3 * 0.44/0.412 =0.32 (32%)
where
P(B∩W)= probability of purchasing the basic model and purchasing the extended warranty
P(B/W) = probability of purchasing the basic model given that has purchased the extended warranty
Final answer:
Using Bayes' theorem, there is approximately a 32% chance that a customer who purchased an extended warranty has a basic model camera.
Explanation:
The question asks us to calculate the probability of a randomly selected purchaser having a basic model given that they have an extended warranty. We can use Bayes' theorem to solve this. Let's denote B as the event of buying a basic model, D as the event of buying a deluxe model, and W as the event of purchasing an extended warranty. From the information provided, we can infer :
P(B) = 0.30 (30% of cameras sold are basic models)
P(D) = 1 - P(B) = 0.70 (since if it's not a basic model, it's a deluxe model)
P(W|B) = 0.44 (44% of basic model buyers purchase an extended warranty)
P(W|D) = 0.40 (40% of deluxe model buyers purchase an extended warranty)
We are interested in P(B|W), the probability that a randomly selected purchaser who bought an extended warranty has a basic model. Using Bayes' theorem:
P(B|W) = (P(W|B) * P(B)) / ((P(W|B) * P(B)) + (P(W|D) * P(D)))
Plugging in the values:
P(B|W) = (0.44 * 0.30) / ((0.44 * 0.30) + (0.40 * 0.70))
P(B|W) = (0.132) / (0.132 + 0.28)
P(B|W) = 0.132 / 0.412 = approximately 0.320
So, there is approximately a 32% chance that a customer with an extended warranty has purchased a basic model camera.
203 red, 117 white, and 28 blue. She asked the students to draw marbles from the box, one at a time and without looking. What is the minimum number of marbles which students should take from the box to ensure that at least three of them are of the same colour?
Answer:
7
Step-by-step explanation:
Since the goal is to draw three marbles of the same colour, regardless of which colour that is, the worst possible scenario would be drawing two marbles of each color in the first six picks (2 red, 2 white and 2 blue). At this point, with the 7th pick, no matter what colour marble the student picks will form three of the same kind.
Therefore, the minimum number of marbles which students should take from the box to ensure that at least three of them are of the same colour is 7.
Final answer:
To ensure at least three marbles of the same color, a minimum of 349 marbles must be drawn from the box.
Explanation:
Question: What is the minimum number of marbles which students should take from the box to ensure that at least three of them are of the same color?
When considering the worst-case scenario, you would need to take marbles until you have at least 3 of the same color.
For this situation, where you have 203 red, 117 white, and 28 blue marbles:
It's possible that you could pick 202 red, 117 white, 27 blue, and still not have 3 of the same color. The next pick would guarantee 3 of the same color, giving a minimum of 349 marbles in total.
"Suppose you draw a single card from a standard deck of 52 cards. How many ways arethere to draw either queen or a heart?"
Answer:
16 ways
Step-by-step explanation:
We are given that
Total cards=52
Total queen in deck of cards=4
Total cards of heart=13
We know 1 card of queen is heart
Number of ways drawing one card of heart out of 13=[tex]13C_1[/tex]
Using combination formula ;[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]
Number of ways drawing one card of heart out of 13=[tex]\frac{13!}{1!12!}=\frac{13\times 12!}{12!}=13[/tex]
Number of ways of drawing one card of queen out of 4=[tex]4C_1=\frac{4!}{3!}[/tex]
Number of ways of drawing one card of queen out of 4=[tex]\frac{4\times 3!}{3!}=4[/tex]
Total number of ways of drawing one card out of 52 cards=[tex]13+4-1=16[/tex]
The accompanying frequency distribution represents the square footage of a random sample of 500 houses that are owner occupied year round. Approximate the mean and standard deviation square footage. statcrunch
Square footage Frequency 0- 499 500 999 13 1,000 1.499 33 1,500 1.999 115 2,000- 2.499 125 2,500 2.999 81 3,000- 3.499 3,500- 3.999 45 4,000 4.499 22 4,500 4.999 10
Answer:
[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{n} = \frac{1220750}{500}=2441.5[/tex]
[tex] s= \sqrt{\frac{N \sum x^2 f -[\sum xf]^2}{N(N-1}}= \sqrt{\frac{500*3408029125 -[1220750]^2}{50*49}}=9341.2405[/tex]
Step-by-step explanation:
In order to find the mean and standard deviation we can create the following table:
Limits Frequency(f) x(midpoint) x*f x^2 *f
__________________________________________________
0-499 9 249.5 2245.5 560252.3
500-999 13 749.5 9743.5 7302753
1000-1499 33 1249.5 41233.5 51521258.25
1500-1999 115 1749.5 201192.5 351986278.8
2000-2499 125 2249.5 281187.5 632531281.3
2500-2999 81 2749.5 222709.5 612339770.3
3000-3499 47 3249.5 152726.5 496284761.8
3500-3999 45 3749.5 168727.5 632643761.3
4000-4499 22 4249.5 93489 397281505.5
4500-4999 10 4749.5 47495 225577502.5
_____________________________________________________
Total 500 1220750 3408029125
We can calculate the mean with the following formula:
[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{n} = \frac{1220750}{500}=2441.5[/tex]
And the standard deviation would be given by:
[tex] s= \sqrt{\frac{N \sum x^2 f -[\sum xf]^2}{N(N-1}}= \sqrt{\frac{500*3408029125 -[1220750]^2}{50*49}}=9341.2405[/tex]
Consider the motion of a free particle, of mass m, described in polar coordinates r, θ. Let its speed be v₀. Let the particle start at t = 0 at r = rᵢₙ, θ = θᵢₙ, and the distance of the closest approach to the origin be r₀.
Show that if the particle is coming in toward the origin (i.e. r is decreasing) that:
dr/dt = -v₀ [ 1 -r₀²/r² ]^(1/2)
Answer:
As proved in the attached files
Step-by-step explanation:
The detailed step and mathematical derivation and manipulation is as shown in the attachment.
The Bagel Club at Middle Township High School sells bagels to teachers for $2 each. Each day they spend $!9 on supplies. How many bagels did the club sell on Thursday if their profit was $55?
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Answer:
32
Step-by-step explanation:
32 X 2 = 64. 64 - 9 = 55.
Find the greatest common factor of the following monomials 32m^5 4m^6
The greatest common factor of 4 and 32 would be 4.
The common factor for m^6 and m^5 would be m^5, since 6 is greater than 5.
The GCF becomes 4m^5