Four (4) decimal places for entire project (when needed) and show what you enter
ID: 3382941 • Letter: F
Question
Four (4) decimal places for entire project (when needed) and show what you entered in the calculator and show picture for partial credit. Use for problems #1 to #4 --- From the data we collected, the population mean height (all my classes) was 68.3 inches with a standard deviation of 3.7 inches. Consider this to be the population. Describe the sampling distribution of the sample mean if we take a sample size of 31 students. Show all three. Find the probability that a randomly selected student will more than 69 inches tall Find the probability that a random sample of 31 students have a mean height of more than 69 inches. Interpret, in a sentence, number 3 above, using the context of the problem.Explanation / Answer
1.
a) It will be approximately normally distributed.
b) It will have a mean of u(X) = 68.3 in.
c) It will have a standard deviation of sigma/sqrt(n) = 3.7/sqrt(31) = 0.664539617.
2.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 69
u = mean = 68.3
s = standard deviation = 3.7
Thus,
z = (x - u) / s = 0.189189189
Thus, using a table/technology, the right tailed area of this is
P(z > 0.189189189 ) = 0.42497227 [ANSWER]
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3.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 69
u = mean = 68.3
n = sample size = 31
s = standard deviation = 3.7
Thus,
z = (x - u) * sqrt(n) / s = 1.053360825
Thus, using a table/technology, the right tailed area of this is
P(z > 1.053360825 ) = 0.146087826 [ANSWER]
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4.
Around 14.61% of the sample means of samples of size 31 will be greater the 69 in.
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