Four 4 decimal places for entire project when needed and sho

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.

Solution

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.