Final answer:
To find the probability that the size of the family is between 2 and 5 inclusive, calculate the proportion of the given sample that falls within that range. In this case, the probability is approximately 0.792.
Explanation:
To find the probability that the size of the family is between 2 and 5 inclusive, we need to calculate the proportion of the given sample that falls within that range. In the provided sample of college math class, the family sizes are:
545443643355633274522232We can see that there are 19 family sizes falling between 2 and 5, inclusive. Therefore, the probability is:
Probability = Number of favorable outcomes / Total number of outcomes
Probability = 19 / 24 = 0.792
So, the probability that the size of the family is between 2 and 5 inclusive is approximately 0.792.
According to a recent study, 23% of U.S. mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of twelve mortgages was selected. What is the probability that greater than 5 of these mortgages are delinquent?
Answer:
P ( X > 5) = 0.0374
Step-by-step explanation:
Given:
n = 12
p = 0.23
Using Binomial distribution formula,
X ~ Binomial ( n = 12, p = 0.23)
[tex]=\frac{n!}{(n-x)! x!}. p^{x} q^{n-x}[/tex]
Substitute for n = 12, p = 0.23, q = 1-0.23 for x = 6,7,8,9,10,11 and 12
P (X > 5) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)
P ( X > 5 ) = 0.0285 + 0.007299 + 0.00136 + 0.000181 + 0.0000162 + 1E-6 + 1E-6
P ( X > 5) = 0.0374
A corporation has 11 manufacturing plants. Of these, seven are domestic and four are outside the United States. Each year a performance evaluation is conducted for four randomly selected plants. What is the probability that a performance evaluation will include at least one plant outside the United States
Answer:
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Step-by-step explanation:
Total plants = 11
Domestic plants = 7
Outside the US plants = 4
Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure
Here we have n=4, p=4/11 and q=7/11
P(X≥1) = 1 - P(X<1)
= 1 - P(X=0)
= 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰
= 1 - 0.16399
P(X≥1) = 0.836
The probability that a performance evaluation will include at least one plant outside the United States is 0.836.
Your large corporation manufactures a certain popular brand of robot lawn mower. After manufacture, all mowers go through a standard quality control check. The Portland factory, which manufactures 30% of your mowers, has a probability of .8 that each mower will pass the quality control check. The Dallas factory, which manufactures 50% of your mowers, has a probability of .7 that each mower will pass the quality control check. For the Buffalo factory,
which manufactures the remaining 20% of your mowers, you have been unable to learn what the probability is that each mower will pass inspection. But, the corporation’s annual report claims that the overall probability that one of their manufactured mowers will pass the quality control check is .72
a) What do you conclude is the probability that a mower manufactured at the Buffalo factory will pass the quality control check?
b) A customer orders one of your mowers, and of course receives one that has passed the quality control check. What is the probability it was manufactured in Dallas?
Answer:
(a) The probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.
(b) The probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.
Step-by-step explanation:
Denote the events as follows:
X = a mower is manufactured at the Portland factory
Y = a mower is manufactured at the Dallas factory
Z= a mower is manufactured at the Buffalo factory
A = a mower passes the quality check.
The information provided is:
[tex]P(X)=0.30\\P(A|X)=0.80\\P(Y)=0.50\\P(A|Y)=0.70\\P(Z)=0.20\\P(A)=0.72[/tex]
(a)
The probability that a mower manufactured at the Buffalo factory will pass the quality control check is:
P (A|Z)
Compute the value of P (A|Z) as follows:
[tex]P(A)=P(A\cap X)+P(A\cap Y) + P (A\cap Z)\\0.72=(0.80\times0.30)+(0.70\times0.50)+(0.20\times P(A|Z))\\0.20\times P(A|Z)=0.72-0.24-0.35\\P(A|Z)=\frac{0.13}{0.20}\\=0.65[/tex]
Thus, the probability that a mower manufactured at the Buffalo factory will pass the quality control check is 0.65.
(b)
Compute the value of P (Y|A) as follows:
[tex]P(Y|A)=\frac{P(A|Y)P(Y)}{P(A)}=\frac{0.70\times0.50}{0.72}=0.4861[/tex]
Thus, the probability that a mower was manufactured in Dallas given that it passes the quality check is 0.4861.
The probability that a mower from the Buffalo factory will pass the quality control check is 13%, and if a customer receives a mower that has passed the check, there is a 49% probability that it was manufactured in Dallas.
Explanation:To find the probability that a robot lawn mower made at the Buffalo factory will pass the quality control check, we first understand that the total probability of a mower passing the check is a sum of all the probabilities from the three factories. This is given as 0.72.
The Portland factory which makes 30% of the mowers has a 0.8 probability of passing the check. Therefore, the contribution of Portland to the total probability is 0.3*0.8 = 0.24. Similarly, the Dallas factory makes 50% of the mowers and each has a passing probability of 0.7, so the Dallas contribution is 0.5*0.7 = 0.35.
Knowing this, we can subtract the total contributions of Portland and Dallas from the overall probability of 0.72 to get Buffalo's contribution, which is the Buffalo passing probability. Therefore, the Buffalo passing probability becomes 0.72 - 0.24 - 0.35 = 0.13 or 13%.
In the second part, if the customer receives a mower that has passed the quality control check, the probability that it was manufactured in Dallas is the contribution of the Dallas factory to the passing mowers i.e., 0.35 ÷ 0.72 = 0.486 or approximately 49%.
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What is the age distribution of adult shoplifters (21 years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution. Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use 45.5 as the class midpoint. (Enter your answers to one decimal place.)
Age range (years) 21-30 31-40 41 and over
Number of shoplifters 280 368 247
We can estimate the mean, variance, and standard deviation by using class midpoints and the number of observations. First, find the midpoint of each age group. Then, to find the mean, multiply each midpoint by the number of shoplifters in that age group, sum those products, and divide by the total number of observations. To find the variance, subtract the mean from each midpoint, square the result, multiply by the number of shoplifters in that group, sum those products, and divide by the total number of observations minus one. The standard deviation is the square root of the variance.
Explanation:To estimate the mean age, sample variance, and sample standard deviation for the shoplifters from a random sample of 895 incidents we need to follow a few steps. Here's how to do it:
Calculate the midpoint of each class: For the age ranges 21-30 and 31-40, the midpoints are 25.5 and 35.5 respectively. The problem already provides 45.5 as the class midpoint for the range '41 and over'.Calculate the estimated mean (µ) by multiplying the midpoint of each class by the number of observations in that class, summing these values, and dividing by the total number of observations.Calculate the estimated variance (σ²) by subtracting the estimated mean from each class midpoint, squaring the result, multiplying by the number of observations in that class, summing these values, and dividing by the total number of observations - 1.Finally, calculate the standard deviation (σ) by taking the square root of the estimated variance.Learn more about Statistics here:https://brainly.com/question/31538429
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14 logs are shipped to a saw mill. The decision must be made on whether they will be classified as Prime, Good, Acceptable, or Not Acceptable. The usual counts are 3 Prime, 5 Good, and 4 Acceptable. In how many ways can these logs be classified so as to match the usual counts
Answer:
2522520
Step-by-step explanation:
Number of ways logs can be classified = 14C5* 9C4*5C3
= 2002*126*10
= 2522520
Number of ways to select 5 good = 14C5, out of remaining 9, number of ways to select acceptable log = 9C4, out of remaining 5, number of ways to select prime log = 5C3 and remaking two unacceptable in 2C2 ways
Final answer:
To find how many ways 14 logs can be classified into categories to match the usual counts, use the multinomial coefficient formula based on the given category counts, resulting in a single calculations. to be 54.6.
Explanation:
The question asks in how many ways 14 logs can be classified into four categories (Prime, Good, Acceptable, Not Acceptable) if we know the usual counts for three of these categories are 3 Prime, 5 Good, 4 Acceptable, and the rest are Not Acceptable. This is a combinatorial problem that can be solved using combinations.
First, we distribute the logs into the Prime, Good, and Acceptable categories as given. This leaves us with 14 - (3 + 5 + 4) = 2 logs to be classified as Not Acceptable. Hence, all of the logs are accounted for with the specified counts.
The number of ways to classify the logs can thus be calculated as the number of ways to choose 3 out of 14 for Prime, then 5 out of the remaining 11 for Good, then 4 out of the remaining 6 for Acceptable. The remaining 2 are automatically classified as Not Acceptable. However, since we're simply fulfilling given counts, and the categories are distinct without overlap, we actually approach this as a partition of 14 objects into parts of fixed sizes, which is a straightforward calculation given by the multinomial coefficient:
The formula for the calculation is: 14! / (3! × 5! × 4! × 2!)
Which simplifies to the total number of ways these logs can be classified according to the specified counts
= 54.6
For a data set of weights (pounds) and highway fuel consumption amounts (mpg) of eight types of automobile, the linear correlation coefficient is found and the P-value is 0.044. Write a statement that interprets the P-value and includes a conclusion about linear correlation.
The P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is [WHAT PERCENT] which is [LOW OR HIGH] so there [IS OR IS NOT] sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.
(Type an integer or a decimal. Do not round.)
Answer:
The P-value indicates that the probability of a linear correlation coefficient that is at least as extreme is 4.4 which is LOW so there IS sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.
Step-by-step explanation:
Hello!
Remember:
The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).
Let's say that the significance level of this correlation test is α: 0.05
If the p-value is the probability of obtaining
you can express it as a percentage: 4.4%Is a very low probability. The decision rule using the p-value is:
p-value < α ⇒ Reject the null hypothesis
p-value ≥ α ⇒ Do not reject the null hypothesis.
The p-value is less than the significance level, the decision is to reject the null hypothesis.
In a linear correlation analysis the statement "there is no linear correlation between the two variables" is always in the null hypothesis, so if you reject it, you can conclude that there is a linear correlation between the variables.
I hope it helps!
In statistical analysis, a P-value of 0.044 indicates there is a 4.4% chance of obtaining a linear correlation as extreme as the observed correlation coefficient. A P-value under 0.05 provides enough evidence to reject the null hypothesis of no correlation, therefore suggesting a significant correlation. Hence, there is sufficient evidence of a linear correlation between car weight and highway fuel consumption.
Explanation:The P-value of 0.044 in this context represents the likelihood of obtaining a linear correlation coefficient for the data points in your dataset that is as extreme as, or more extreme than, the one you calculated, assuming there is no linear relationship between the two variables (weight and highway fuel consumption in automobiles). This probability is 4.4% - rather low. A common threshold for significance in many fields is 0.05, or 5%. If your P-value is below this threshold, we reject the null hypothesis that there is no correlation and conclude there may be a correlation. Therefore, as the P-value is 0.044, which is below the 0.05 threshold, there is sufficient evidence to conclude that there is a linear correlation between weight and highway fuel consumption in automobiles.
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Professor stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particulare on the the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, also independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half of the crossings encountered.
A.) Which route should he take to minimize the probability of being late to the meeting?
B.) If he tosses a fair coin to decide on a route and is late, what is the probability he took the four crossing route?
Answer:
A) He should take the route with 4 crossings
B) The probability that he took the 4 crossing route is 0.2158
Step-by-step explanation:
Lets call X the number of crossings he encounters, A if he takes route 1 and B if he takes route 2.
Note that X given A is a binomial random variable with parameters n = 4 p = 0.1, and X given B has parameters n = 2, p = 0.1
The probability that the Professor is on time on route 1 is equal to
P(X|A = 0) + P(X|A = 1) = 0.9⁴ + 0.9³*0.1*4 = 0.9477
On the other hand, the probability that the professor is on time on route 2 is
P(X|B = 0) = 0.9² = 0.81
Hence, it is more likely for the professor to be late on route 2, thus he should take the route 1, the one with 4 crossings.
B) Lets call L the event 'The professor is late'. We know that
P(L|A) = 1-0.9477 = 0.0524
P(L|B) = 1-0.81 = 0.19
Also
P(A) = P(B) = 1/2 (this only depends on the result of the coin.
For the Bayes theorem we know, therefore that
[tex]P(A|L) = \frac{P(L|A) * P(A)}{P(L|A)*P(A) + P(L|B)*P(B)} = \frac{0.0523*0.5}{0.0523*0.5 + 0.19*0.5 } = 0.2158[/tex]
Hence, the probability that he took the 4 crossing route is 0.2158.
Your body loses sodium when you sweat. Researchers sampled 38 random tennis players. The average sodium loss was 500 milligrams per pound and the standard deviation was 62 milligrams per pound. Construct and interpret a 99% confidence interval to estimate the mean loss in sodium in the population.
Answer:
The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.
Step-by-step explanation:
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 M 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.
[tex]M = 2.575*\frac{62}{\sqrt{38}} = 25.90[/tex]
The lower end of the interval is the mean subtracted by M. So it is 500 - 25.90 = 474.10 milligrams.
The upper end of the interval is the mean added to M. So it is 500 + 25.90 = 525.90 milligrams
The 99% confidence interval to estimate the mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams. This means that we are 99% that the true mean loss in sodium in the population is between 474.10 milligrams and 525.90 milligrams.
Answer:
Option A
Step-by-step explanation:
The 99% confidence interval is (472.69, 527.31). We are 99% confident that the true population mean of sodium loss for tennis players will be between 472.69 milligrams per pound and 527.31 milligrams per pound.
Can someone please help me find the missing lengths in the following diagram?
Answer:
Step-by-step explanation:
18 a)Since line DE is parallel to line BC, it means that triangle ADE is similar to triangle ABC. Therefore,
AB/AD = AC/AE = BC/DE
AC = AE + CE = 18 + 9
AC = 27
AB = AD + DB
AB = AD + 5
Therefore,
27/18 = (AD + 5)/AD
Cross multiplying, it becomes
27 × AD = 18(AD + 5)
27AD = 18AD + 90
27AD - 18AD = 90
9AD = 90
AD = 90/9
AD = 10
b) AC = AE + EC = 13 + 3
AC = 16
To find AD,
16/13 = 24/AD
16 × AD = 13 × 24
16AD = 312
AD = 312/16
AD = 19.5
DB = 24 - 19.5
DB = 4.5
Data collected at Toronto Pearson International Airport suggest that an exponential distribution with mean value 2,725 hours is a good model for rainfall duration. What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?
Answer:
[tex] P(X >2) [/tex]
And we can calculate this with the complement rule like this:
[tex] P(X>2) = 1-P(X<2)[/tex]
And using the cdf we got:
[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]
Step-by-step explanation:
Previous concepts
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}, x>0[/tex]
And 0 for other case. Let X the random variable of interest:
[tex]X \sim Exp(\lambda=\frac{1}{2.725})[/tex]
Solution to the problem
We want to calculate this probability:
[tex] P(X >2) [/tex]
And we can calculate this with the complement rule like this:
[tex] P(X>2) = 1-P(X<2)[/tex]
And using the cdf we got:
[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]
Suppose X is a normal distribution with N(210, 32). Find the following: a. P( X < 230) b. P(180 < X < 245) c. P( X >190) d. Find c such that P( X < c) = 0.0344 e. Find c such that P( X > c) = 0.7486
Using the normal distribution, it is found that:
a) P(X < 230) = 0.734.
b) P(180 < X < 245) = 0.6885.
c) P( X >190) = 0.734.
d) X = 151.76.
e) X = 188.56.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It 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, which is the percentile of X.In this problem:
The mean is of [tex]\mu = 210[/tex].The standard deviation is of [tex]\sigma = 32[/tex].Item a:
This probability is the p-value of Z when X = 230, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{230 - 210}{32}[/tex]
[tex]Z = 0.625[/tex]
[tex]Z = 0.625[/tex] has a p-value of 0.734.
Hence:
P(X < 230) = 0.734.
Item b:
This probability is the p-value of Z when X = 245 subtracted by the p-value of Z when X = 180, hence:
X = 245
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{245 - 210}{32}[/tex]
[tex]Z = 1.09[/tex]
[tex]Z = 1.09[/tex] has a p-value of 0.8621.
X = 180
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{180 - 210}{32}[/tex]
[tex]Z = -0.94[/tex]
[tex]Z = -0.94[/tex] has a p-value of 0.1736.
0.8621 - 0.1736 = 0.6885.
Then:
P(180 < X < 245) = 0.6885.
Item c:
This probability is 1 subtracted by the p-value of Z when X = 190, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{190 - 210}{32}[/tex]
[tex]Z = -0.625[/tex]
[tex]Z = -0.625[/tex] has a p-value of 0.266.
1 - 0.266 = 0.734.
Hence:
P( X >190) = 0.734.
Item d:
This is X = c when Z has a p-value of 0.0344, hence X when Z = -1.82.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.82 = \frac{X - 210}{32}[/tex]
[tex]X - 210 = -1.82(32)[/tex]
[tex]X = 151.76[/tex]
Item e:
This is X when Z has a p-value of 1 - 0.7486 = 0.2514, hence X when Z = -0.67.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.67 = \frac{X - 210}{32}[/tex]
[tex]X - 210 = -0.67(32)[/tex]
[tex]X = 188.56[/tex]
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The question involves calculating probabilities and specific values (c) for a given normal distribution with mean 210 and standard deviation 32. The probabilities for certain ranges and tail ends are computed using the normal cumulative distribution function and its inverse.
Explanation:The question pertains to finding probabilities and specific values associated with a normal distribution X which is denoted as N(210, 32), meaning it has a mean (μ) of 210 and a standard deviation (σ) of 32.
P(X < 230) can be found using a z-score calculation or a normal cumulative distribution function. Since this is a left-tail probability, we're interested in the area under the curve to the left of X=230.P(180 < X < 245) is the probability that X falls between these two values. We look at the area under the normal curve between these two points.P(X > 190) represents the right-tail probability, meaning the area under the curve to the right of X=190.To find c such that P(X < c) = 0.0344, we'd use the inverse of the normal cumulative distribution function, often denoted as the quantile or the percentile function.Finding the value c such that P(X > c) = 0.7486 again involves the use of the inverse normal function, but this time looking at the right-tail probability.On a test with a population mean of 75 and standard deviation equal to 16, if the scores are normally distributed, what percentage of scores fall between 70 and 80?
Answer:
Percentage of scores that fall between 70 and 80 = 24.34%
Step-by-step explanation:
We are given a test with a population mean of 75 and standard deviation equal to 16.
Let X = Percentage of scores
Since, X ~ N([tex]\mu,\sigma^{2}[/tex])
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1) where, [tex]\mu[/tex] = 75 and [tex]\sigma[/tex] = 16
So, P(70 < X < 80) = P(X < 80) - P(X <= 70)
P(X < 80) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{80-75}{16}[/tex] ) = P(Z < 0.31) = 0.62172
P(X <= 70) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{70-75}{16}[/tex] ) = P(Z < -0.31) = 1 - P(Z <= 0.31)
= 1 - 0.62172 = 0.37828
Therefore, P(70 < X < 80) = 0.62172 - 0.37828 = 0.24344 or 24.34%
Suppose that the waiting time for an elevator at a local shopping mall is uniformly distributed from 0 to 90 seconds.
What is the probability that a customer waits for more than 60 seconds?
Answer:
1/3
Step-by-step explanation:
60-90 is 30 numbers, right? So it is 30/90, or 1/3
Which of the following are continuous variables, and which are discrete? (a) number of traffic fatalities per year in the state of Florida continuous discrete (b) distance a golf ball travels after being hit with a driver continuous discrete (c) time required to drive from home to college on any given day continuous discrete (d) number of ships in Pearl Harbor on any given day continuous discrete (e) your weight before breakfast each morning continuous discrete
Answer:
a) Discrete
b) Continuous
c) Continuous
d) Discrete
e) Continous
Step-by-step explanation:
Continuous:
Real numbers, can be integer, decimal, etc.
Discrete:
Only integer(countable values). So can be 0,1,2...
(a) number of traffic fatalities per year in the state of Florida
You cannot have half of a traffic fatality, for example. So this is discrete
(b) distance a golf ball travels after being hit with a driver
The ball can travel 10.25m, for example, which is a decimal number. So this is continuous.
(c) time required to drive from home to college on any given day
You can take 10.5 minutes, for example, which is a decimal value. So this is continuous.
(d) number of ships in Pearl Harbor on any given day
There is no half ship, for example. So this is discrete.
(e) your weight before breakfast each morning continuous discrete
You can weigh 80.4kg, for example, which is a decimal number. So this is continuous.
a) Discrete
b) Continuous
c) Continuous
d) Discrete
e) Continuous
Discrete data is the numerical type of data which includes whole, concrete numbers that has specific and fixed data values. therefore, they can be determined by counting.
example: Number of students in the class, ect.,
Continuous type of data includes complex numbers or the varying data values that are measured over a specific time interval.
example: Height of students in a school, etc.,
(a) Number of traffic fatalities per year in the state of Florida.
This is a Discrete data, as it is a countable data and will be always a whole be number. Number of traffic fatalities per year will be 75, 150, 200, etc.
(b) distance a golf ball travels after being hit with a driver.
This is a Discrete data, as the data can be measures and will be not always be a whole number. distance a golf ball travels after being hit can be 1.25 meters, 10.9 meters, 21.1 meters, etc.
(c) time required to drive from home to college on any given day.
This is a Discrete data, as the data can be measures and will be not always be a whole number. time required to drive from home to college can be 5minutes 30 seconds, 15 minutes 26 seconds.
(d) number of ships in Pearl Harbor on any given day.
This is a Discrete data, as it is a countable data and will be always a whole number. number of ships in Pearl Harbor on any given day will be always 1, 10, 8, etc.
(e) your weight before breakfast each morning.
This is a Discrete data, as the data can be measures and will be not always be a whole number. weight can be 44.56 kgs., 52.3 kgs, etc.
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The lengths of the sides of a rectangle are consecutive prime numbers. The area of the rectangle is represented by a three-digit number composed only of the two smallest prime digits and it will not change if we reverse it . What is the perimeter of the rectangle?
WILL MARK BRAINLIEST!
Answer:
Perimeter = 72
Step-by-step explanation:
Mathematics Puzzle
The area of a rectangle is the product of the base and the height
A=b.h
We know the base and the height are two consecutive prime numbers, not much of useful information so far.
We also know the area is composed only of the two smallest prime digits. Those digits are 2 and 3. If the number is reversed and it's not changed, then we only have two possible values for the area: 232 and 323.
We only need to find two consecutive prime numbers which product is one of the above. The number 232 has no prime factors: 232 = 8*29.
The number 323 is the product of 17 and 19, two consecutive prime numbers, thus the dimensions of the rectangle are 17 and 19.
The perimeter of that rectangle is 2*17+2*19= 72
Answer:
72 units
Step-by-step explanation:
72 units
What is the product?
Answer: the third option is correct
Step-by-step explanation:
The first matrix is a 2 × 3 matrix while the second matrix is a 3 × 2 matrix. To get the product of both matrices, we would multiply each term in each row by the terms in the corresponding column and add.
1) row 1, column 1
1×2 + 3×3 + 1×4 = 2 + 9 + 4 = 15
2) row 1, column 2
1×-2 + 3×5 + 1×1 = - 2 + 15 + 1 = 14
3) row 2, column 1
-2×-2 + 1×3 + 0×4 = - 4 + 3 + 0 = - 1
4) row 2, column 2
- 2×-2 + 1×5 + 0×1 = 4 + 5 = 9
The solution becomes
15 14
- 1 9
1. You have an aluminum bar of dimensions 2cm*5cm*10 cm. You want to put it into electric circuit such a way that this bar will demonstrate the smallest possible resistance. You should connect your bar to the opposite faces with dimensions of:
Answer:
( 5 x 10 ) cm
Step-by-step explanation:
Given:
- The dimensions of the bar are:
( 2 x 5 x 10 ) cm
Find:
Which two faces with dimensions ( _x _ ) should be connected to get smallest possible resistance.
Solution:
- The electrical resistance R of any material with density ρ and corresponding dimensions is expressed as:
R = ρ*L / A
- Where, A: cross sectional Area
L: The length in between the two faces.
- We need to minimize the electrical resistance of the bar. For that the Area must be maximized and Length should be minimized.
- A_max & L_min ---- > R_min
(5*10) & ( 2) ------> R_min
Hence, the electrical resistance is minimized by connecting the face with following dimensions ( 5 x 10 ) cm
Here are summary statistics for randomly selected weights of newborn girls: nequals202, x overbarequals28.3 hg, sequals6.1 hg. Construct a confidence interval estimate of the mean. Use a 95% confidence level. Are these results very different from the confidence interval 27.8 hgless thanmuless than29.6 hg with only 17 sample values, x overbarequals28.7 hg, and sequals1.8 hg? What is the confidence interval for the population mean mu? nothing hgless thanmuless than nothing hg (Round to one decimal place as needed.) Are the results between the two confidence intervals very different? A. Yes, because the confidence interval limits are not similar. B. Yes, because one confidence interval does not contain the mean of the other confidence interval. C. No, because each confidence interval contains the mean of the other confidence interval. D. No, because the confidence interval limits are similar.
Answer:
The confidence interval is 27.5 hg less than mu less than 29.1 hg
(A) Yes, because the confidence interval limits are not similar.
Step-by-step explanation:
Confidence interval is given as mean +/- margin of error (E)
mean = 28.3 hg
sd = 6.1 hg
n = 202
degree of freedom = n-1 = 202-1 = 201
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196
E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg
Lower limit = mean - E = 28.3 0.8 = 27.5 hg
Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg
95% confidence interval is (27.5, 29.1)
When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17
The confidence intervals for the two experiments have overlapping intervals, suggesting that the results are not very different.
Explanation:The first experiment resulted in a 95% confidence interval of 3.070-3.164 g for the population mean weight of newborn girls. The second experiment had a 95% confidence interval of 3.035-3.127 g. Although the two confidence intervals are not identical, the mean for each experiment is within the confidence interval of the other experiment. This suggests that the results are not very different and there is an appreciable overlap between the two intervals. Therefore, the answer is C. No, because each confidence interval contains the mean of the other confidence interval.
In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2. Section 05.01 Exercise 12.a - Compute confidence interval; Find necessary sample size Find a 95% confidence interval for the mean efficiency. Round the answers to three decimal places.
Answer:
95% confidence interval for the mean efficiency is [84.483 , 85.517].
Step-by-step explanation:
We are given that in a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.
So, the pivotal quantity for 95% confidence interval for the population mean efficiency is given by;
P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\mu[/tex] = sample average efficiency = 85
[tex]\sigma[/tex] = sample standard deviation = 2
n = sample of motors = 60
[tex]\mu[/tex] = population mean efficiency
So, 95% confidence interval for the mean efficiency, [tex]\mu[/tex] is ;
P(-2.0009 < [tex]t_5_9[/tex] < 2.0009) = 0.95
P(-2.0009 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.0009 ) = 0.95
P( [tex]-2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P( [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.0009 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +2.0009 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]85 -2.0009 \times {\frac{2}{\sqrt{60} }[/tex] , [tex]85 +2.0009 \times {\frac{2}{\sqrt{60} }[/tex] ]
= [84.483 , 85.517]
Therefore, 95% confidence interval for the population mean efficiency is [84.483 , 85.517].
Final answer:
The 95% confidence interval for the mean efficiency of electric motors, given a sample mean of 85, a standard deviation of 2, and a sample size of 60, is approximately (84.494, 85.506) when rounded to three decimal places.
Explanation:
To calculate the 95% confidence interval for the mean efficiency of electric motors, we use the sample mean, standard deviation, and the sample size along with the z-score for the 95% confidence level. Since the sample size is large (n > 30), we can use the z-distribution to approximate the sampling distribution of the sample mean.
The formula for a confidence interval is:
Confidence Interval = sample mean "+/-" (z-score * (standard deviation / sqrt(n)))
Given that the sample mean is 85, the standard deviation is 2, and the sample size (n) is 60, and using the z-score of approximately 1.96 for 95% confidence, we can compute the confidence interval.
Confidence Interval = 85 "+/-" (1.96 * (2 / √(60)))
After the calculation, we get the two ends of the interval:
Lower Limit = 85 - (1.96 * (2 / √(60))) = 84.494
Upper Limit = 85 + (1.96 * (2 / √(60))) = 85.506
Rounding these to three decimal places, the 95% confidence interval for the mean efficiency of electric motors is approximately (84.494, 85.506). This means we can be 95% confident that the true average efficiency of all electric motors is between 84.494% and 85.506%.
A map uses the scale 1.5 cm = 25 mi. Two cities are 190 miles apart. How far apart are the cities on the map?
Please answer quickly and I will give brainiest to whoever is correct fastest.
Use simple unitary method:
∵ 1.5 cm on map = 25 miles in reality
∴ x cm on map = 190 miles in reality
[tex]\frac{1.5}{x} = \frac{25}{190}\\ x = 11.4 cm[/tex]
Thus, the two cities are 11.4 cm apart on the map
Final answer:
To determine the map distance between two cities that are 190 miles apart using the scale 1.5 cm = 25 miles, divide the actual distance by the scale ratio (16.67 miles/cm), resulting in 11.4 cm.
Explanation:
For the map with scale 1.5 cm = 25 miles, we first calculate how many miles one centimeter represents by dividing the miles by the centimeters in the scale:
25 miles / 1.5 cm = 16.67 miles/cm
Next, we find the map distance for 190 miles by dividing the actual distance by the distance one centimeter represents:
190 miles / 16.67 miles/cm = 11.4 cm
So, the two cities are 11.4 cm apart on the map.
The graph shows the relationship between the number of months different students practiced baseball and the number of games they won:
The title of the graph is Baseball Games. On x axis, the label is Number of Months of Practice. On y axis, the label is Number of Games Won. The scale on the y axis is from 0 to 22 at increments of 2, and the scale on the x axis is from 0 to 12 at increments of 2. The points plotted on the graph are the ordered pairs 0, 1 and 1, 3 and 2, 5 and 3, 9 and 4, 10 and 5, 12 and 6, 13 and 7, 14 and 8,17 and 9, 18 and 10,20. A straight line is drawn joining the ordered pairs 0, 1.8 and 2, 5.6 and 4, 9.2 and 6, 13 and 8, 16.5 and 10, 20.5.
Part A: What is the approximate y-intercept of the line of best fit and what does it represent? (5 points)
Part B: Write the equation for the line of best fit in slope-intercept form and use it to predict the number of games that could be won after 13 months of practice. Show your work and include the points used to calculate the slope.
Answer:
A) from the line of best fit, the approximately y-intercept is (0,1.8). This means without any practice, 1h.8 games are won.
B) slope: (5.6-1.8)/(2-0) = 1.9
y = 1.9x + 1.8
(Line of best fit)
x = 13,
y = 1.9(13) + 1.8 = 26.5
Predicted no. of games won after 13 months of practice is 26.5
Final answer:
The y-intercept, representing initial games won and the equation for line of best fit predicting future games won, are determined from the graph data.
Explanation:
Part A:
The y-intercept of the line of best fit is approximately 1.8.It represents the initial number of games won when the number of months of practice is zero.Part B:
The equation for the line of best fit in slope-intercept form is y = 1.4x + 1.8.
To predict the number of games won after 13 months of practice, substitute x = 13 into the equation:
y = 1.4(13) + 1.8 = 19.5
So, the predicted number of games that could be won after 13 months of practice is 19.5.
derived the MOM and MLE for an exponential distribution with parameter ????. Conduct a Bootstrap simulation to compare the estimation of λ with sample sizes of n = 10, n = 100, and n = 500. Choose true value λ = 0.2 and use B = 1000. Calculate and compare the mean and standard error for each set of simulations to each other as well as their theoretical values.
Answer:
rm(list=ls(all=TRUE))
set.seed(12345)
N=c(10,100,500)
Rate=0.2
B=1000
MN=SE=rep()
for(i in 1:length(N))
{
n=N[i]
X=rexp(n,rate=Rate)
EST=1/mean(X)
ESTh=rep()
for(j in 1:B)
{
Xh=rexp(n,rate=EST)
ESTh[j]=1/mean(Xh)
}
MN[i]=mean(ESTh)
SE[i]=sd(ESTh)
}
cbind(N,Rate,MN,SE)
A certain vibrating system satisfies the equation u'' + γu' + u = 0. Find the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.
In damped harmonic motion, we calculate damping coefficient γ by comparing the periods of damped and undamped motion. For the given situation where the quasi-period is 90% greater than the undamped period, the damping coefficient is approximately 0.7416.
Explanation:The subject of this question involves Damped Harmonic Motion, a concept in Physics, related to vibrations and waves. The equation given, u'' + γu' + u = 0, describes the motion where γ denotes the damping coefficient. Here, we have to calculate this damping coefficient when the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion.
To solve this, we must use the relationship between damped and undamped periods. The quasi-period T' of a damped harmonic motion relates to the undamped period T as: T' = T/(sqrt(1 - (γ/2)^2)). Now, given that T' = 1.9T, we can but these two equations together:
1.9 = 1/(sqrt(1 - (γ/2)^2))
Solving this for γ, we get γ ≈ 0.7416. Hence, the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is approximately 0.7416.
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The value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the undamped motion is the one that satisfies γ=2*ω*0.9, where ω is the natural frequency of oscillation.
Explanation:The given equation is for a damped harmonic oscillator, a physical system that oscillates under both a restoring force and a damping force proportional to the velocity of the system. The damping coefficient γ determines the behavior of the system and in this case, we need to find the value of γ such that the quasi period of the damped motion is 90% greater than the period of the undamped motion.
The period of the undamped motion, T₀, is calculated by the formula T₀=2π/sqrt(ω), where ω is the natural frequency of oscillation. The quasi period of the damped motion, Td, is increased by a factor of 1+η (in this case, 1.9 as the increase is 90%) and calculated by the formula Td=T₀(1+η) = T₀*1.9.
The damping ratio η is determined by the damping coefficient γ as η=γ/2ω. Therefore, by combining these expressions and rearranging the terms, we extract γ from these formulas as γ=2ω*η => γ=2*ω*(0.9). Thus, the value of the damping coefficient γ for which the quasi period of the damped motion is 90% greater than the period of the corresponding undamped motion is the one which satisfies γ=2*ω*0.9.
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Mrs. Porcelli's classroom bulletin board is 2 % feet long. Ms. Smith's bulletin board is 3
| times as long as Mrs. Porcelli's. How long is Ms. Smith's bulletin board.
Answer: 0.06 ft long
Step-by-step explanation:
Porcelli's board = 2% feet long; to convert 2% into fraction, divide by 100= 2/100 = 0.02 ft
Smith's board, from the question is 3 times porcelli's board = 3 x 0.02= 0.06 ft
I hope this helps.
QUESTION 7 A Randstad/Harris interactive survey reported that 25% of employees said their company is loyal to them. Suppose 9 employees are selected randomly and will be interviewed about company loyalty. What is the probability that none of the 9 employees will say their company is loyal to them? g
Answer:
7.51% probability that none of the 9 employees will say their company is loyal to them.
Step-by-step explanation:
For each employee, there are only two possible outcomes. Either they think that their company is loyal to them, or they do not think this. The probability of an employee thinking that their company is loyal to them is independent of other employees. 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.
25% of employees said their company is loyal to them.
This means that [tex]p = 0.25[/tex]
9 employees are selected randomly
This means that [tex]n = 9[/tex]
What is the probability that none of the 9 employees will say their company is loyal to them?
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.25)^{0}.(0.75)^{9} = 0.0751[/tex]
7.51% probability that none of the 9 employees will say their company is loyal to them.
A tennis club offers two payment options:
Option1: $35 monthly fee plus $4/hour for court rental
Option 2: No monthly fee but $6.50/hour for court rental.
Let x = hours per month of court rental time.
a) Write a mathematical model representing the total monthly cost, C, in terms of x for the following:
Option 1: C= _________________
Option 2: C=_________________
b) How many hours would you have to rent the court so that the monthly cost of option 1, is less than option 2. Set up an inequality and show your work algebraically using the information in part a.
Multiply the cost per hour by number of hours x and for option 1 you need to add the monthly fee:
A) option 1: C= 4x +35
Option2: C = 6.50x
B)
4x+35 < 6.50x
Subtract 4x from both sides:
35 < 2.50x
Divide both sides by 2.50 :
X < 14
You would need to rent more than 14 hours for option 1 to be cheaper.
These data can be approximated quite well by a N(3.4, 3.1) model. Economists become alarmed when productivity decreases. According to the normal model what is the probability that the percent change in worker output per hour from the previous quarter is more than 0.6 standard deviations below the mean? .0228 Incorrect: Your answer is incorrect. Question 3. What is the probability that the percent change in worker output from the previous quarter is between -1.715 and 7.12? Use the normal model mentioned at the beginning of question 2.
Answer:
First part
[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]
And for this case we can use the z score formula given by:
[tex] z = \frac{x- \mu}{\sigma}[/tex]
And using this formula we got:
[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]
And we can use the normal standard table or excel and we got:
[tex]P(Z<-0.6) = 0.274[/tex]
Second part
For the other part of the question we want to find the following probability:
[tex] P(-1.715 <X< 7.12)[/tex]
And using the score we got:
[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]
And we can find this probability with this difference:
[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/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 data of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(3.4,3.1)[/tex]
Where [tex]\mu=3.4[/tex] and [tex]\sigma=3.1[/tex]
First part
And for this case we want this probability:
[tex] P(X< 3.4-0.6*3.1) = P(X<1.54)[/tex]
And for this case we can use the z score formula given by:
[tex] z = \frac{x- \mu}{\sigma}[/tex]
And using this formula we got:
[tex] P(X<1.54) = P(Z<\frac{1.54 -3.4}{3.1})= P(Z<-0.6)[/tex]
And we can use the normal standard table or excel and we got:
[tex]P(Z<-0.6) = 0.274[/tex]
Second part
For the other part of the question we want to find the following probability:
[tex] P(-1.715 <X< 7.12)[/tex]
And using the score we got:
[tex] P(-1.715 <X< 7.12)=P(\frac{-1.715-3.4}{3.1} < Z< \frac{7.15-3.4}{3.1}) = P(-1.65< Z< 1.210)[/tex]
And we can find this probability with this difference:
[tex]P(-1.65< Z< 1.210)=P(Z<1.210)-P(z<-1.65) = 0.887-0.049=0.837[/tex]
Given an acceleration vector, initial velocity left angle u 0 comma v 0 right angle, and initial position left angle x 0 comma y 0 right angle, find the velocity and position vectors for tgreater than or equals0. a(t)equalsleft angle 0 comma 12 right angle, left angle u 0 comma v 0 right angleequalsleft angle 0 comma 6 right angle, left angle x 0 comma y 0 right angleequalsleft angle 6 comma negative 1 right angle
Answer:
For the velocity vector, we have
V(t) = dR
dt = (1, 2t, 3t
2
).
For the acceleration vector, we get
A =
dV
dt = (0, 2, 6t).
The velocity vector at t = 1 is
V(1) = (1, 2, 3).
The speed at t = 1 is
kV(1)k =
p
1
2 + 22 + 32 =
√
14.
19) If an average of 12 customers are served per hour, what is the probability that the next customer will arrive in 3 minutes or less? Note: λ = 12/60
Answer:
The probability that the next customer will arrive in 3 minutes or less is 0.45.
Step-by-step explanation:
Let N (t) be a Poisson process with arrival rate λ. If X is the time of the next arrival then,
[tex]P(X>t)=e^{-\lambda t}[/tex]
Given:
[tex]\lambda=\frac{12}{60}[/tex]
t = 3 minutes
Compute the probability that the next customer will arrive in 3 minutes or less as follows:
P (X ≤ 3) = 1 - P (X > 3)
[tex]=1-e^{-\frac{12}{60}\times3}\\=1-e^{-0.6}\\=1-0.55\\=0.45[/tex]
Thus, the probability that the next customer will arrive in 3 minutes or less is 0.45.
The probability that the next customer will arrive in 3 minutes or less is; 0.4512
This is a Poisson distribution problem with the formula;
P(X > t) = e^(-λt)
Where;
λ is arrival rate
t is arrival time
We are given;
λ = 12/60
t = 3 minutes
We want to find the probability that the next customer will arrive in 3 minutes or less. This is expressed as;P (X ≤ 3) = 1 - (P(X > 3))
Thus;
P (X ≤ 3) = 1 - e^((12/60) × 3)
P (X ≤ 3) = 1 - 0.5488
P (X ≤ 3) = 0.4512
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slove this problem find the value of x.
Answer:
x = 6 units.
Step-by-step explanation:
By Geometric mean property:
[tex]x = \sqrt{3 \times 12} = \sqrt{36} = 6 \\ \hspace{20 pt} \huge \orange{ \boxed{ \therefore \: x = 6}}[/tex]
Hence, x = 6 units.