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2. The example on page 5 of chapter 6 notes asks for the probability for a single and (part a) and then for the probability for a mean of a sample of siz n = 7 (part b). Which option includes the two correct probabilities correctly rounded for this example? O P(x>72) = 0.1586, and P(x>72) = 0.0040 P(x >72) = 0.0040, and P(x>72) = 0.1586 P(x>72) = 0.0041, and P(x>72) = 0.1587 P(x >72) = 0.1587, and P(x>72) = 0.0041

2. The example on page 5 of chapter 6 notes asks for the probability for a single and (part a) and then for the probability for a mean of a sample of siz n = 7 (part b). Which option includes the two correct probabilities correctly rounded for this example? O P(x>72) = 0.1586, and P(x>72) = 0.0040 P(x >72) = 0.0040, and P(x>72) = 0.1586 P(x>72) = 0.0041, and P(x>72) = 0.1587 P(x >72) = 0.1587, and P(x>72) = 0.0041

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Hi, I added a picture of the notes in reference to the question. I appreciate the help, thanks

2. The example on page 5 of chapter 6 notes asks for the probability for a single and (part a) and then for the probability for a mean of a sample of size n = 7 (part b). Which option includes the two correct probabilities correctly rounded for this example? OP(x >72) = 0.1586, and P(x>72) = 0.0040 OP(x >72) = 0.0040, and P(x>72) = 0.1586 P(x>72) = 0.0041, and P(x>72) = 0.1587 P(x >72) = 0.1587, and P(x>72) = 0.0041
Transcribed Image Text:2. The example on page 5 of chapter 6 notes asks for the probability for a single and (part a) and then for the probability for a mean of a sample of size n = 7 (part b). Which option includes the two correct probabilities correctly rounded for this example? OP(x >72) = 0.1586, and P(x>72) = 0.0040 OP(x >72) = 0.0040, and P(x>72) = 0.1586 P(x>72) = 0.0041, and P(x>72) = 0.1587 P(x >72) = 0.1587, and P(x>72) = 0.0041
We demonstrated previously how the tossing of a coin (counting results) can have a sampling distribution. It is possible to go through a similar process with the means of measured data. For example, we had a sample of striped dolphins of size n = 20 and that sample had a sample mean of weights. But if we took several different samples of size n = 20 of striped dolphins weights, each sample would have a slightly different sample mean. which is Use the idea of sampling distributions for means, we get the CLT for means, not for counting, it is for measuring. In other words, when we are taking samples of measurements (instead of counting) we still get a sampling distribution with some samples with means below average and some samples with means above average. 4 We still need to have the samples be large enough for the distribution to be symmetric and have a normal distribution. Central Limit Theorem for the Sampling Distribution of the Mean (CLT) The central limit theorem for means (CLT) states that given a sufficiently large sample size (n> 30) from a population: (1) The mean of all samples from the same population will be approximately equal to the mean of the population. (2) All of the samples will follow an approximate normal distribution pattern. (3) All variances will be approximately equal to the variance of the population divided by each sample's size. The notation and formulas for the CLT are: population standard deviation sample standard deviation = S S standard deviation among the samples = = √/₁ Example The mean height of men is 69 inches with a standard deviation of 3 inches. Find (a) The probability of selecting one man who is over 72 inches tall. (This is not the mean of a sample so we will not use the CLT.) = 0 (b) The probability of selecting a group of 7 men and the group mean is over 72 inches. (This is the mean of a sample, so we will use the CLT.)
Transcribed Image Text:We demonstrated previously how the tossing of a coin (counting results) can have a sampling distribution. It is possible to go through a similar process with the means of measured data. For example, we had a sample of striped dolphins of size n = 20 and that sample had a sample mean of weights. But if we took several different samples of size n = 20 of striped dolphins weights, each sample would have a slightly different sample mean. which is Use the idea of sampling distributions for means, we get the CLT for means, not for counting, it is for measuring. In other words, when we are taking samples of measurements (instead of counting) we still get a sampling distribution with some samples with means below average and some samples with means above average. 4 We still need to have the samples be large enough for the distribution to be symmetric and have a normal distribution. Central Limit Theorem for the Sampling Distribution of the Mean (CLT) The central limit theorem for means (CLT) states that given a sufficiently large sample size (n> 30) from a population: (1) The mean of all samples from the same population will be approximately equal to the mean of the population. (2) All of the samples will follow an approximate normal distribution pattern. (3) All variances will be approximately equal to the variance of the population divided by each sample's size. The notation and formulas for the CLT are: population standard deviation sample standard deviation = S S standard deviation among the samples = = √/₁ Example The mean height of men is 69 inches with a standard deviation of 3 inches. Find (a) The probability of selecting one man who is over 72 inches tall. (This is not the mean of a sample so we will not use the CLT.) = 0 (b) The probability of selecting a group of 7 men and the group mean is over 72 inches. (This is the mean of a sample, so we will use the CLT.)
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