Question

You will be performing an analysis on female heights, given of set of 30 heights that were randomly obtained. For this project, it is necessary to know that the average height for women is assumed to be 65 inches with a standard deviation of 3.5 inches. You will use these numbers in some of your calculations. 2. Find Elizabeth in the data and, given the population mean and standard deviation, calculate: a. The z-score for Elizabeth (easiest to be done by hand). b. The probability that a randomly selected female is shorter than Elizabeth. c. The probability that a randomly selected female is taller than Elizabeth. d. Interpret each of these in a sentence (3 full sentences). Mean is 65 inches Standard Deviation is 3.5 inches Elizabeth's height is 65.40 inches 4. In StatCrunch or Excel, find and state the mean of the 30 women’s heights that were provided (round to two decimal places). Then (using the values from step 3), calculate the probability that a random sample of 30 women’s heights would result in a mean that was the value you found or more. Use this probability in a full sentence.

5. Explain how what you just calculated in step 4 relates to the question you were asked in step 2c. Why is the probability in step 4 lower than that in step 2?

Name 72.44 Emma 67.53 Olivia 66.71 Ava 62.02 Isabella 73.89 Sophia 65.95 Mia 65.83 Charlotte 64.15 Amelia 65.39 Evelyn 59.68 Abigail 64.24 Harper 66.60 Emily 65.40 Elizabeth 64.72 Avery 67.11 Sofia 61.97 Ella 62.83 Madison 67.20 Scarlett 66.62 Victoria 68.78 Aria 66.13 Grace 64.47 Chloe 66.64 Camila 62.39 Penelope 63.90 Riley 62.97 Layla 59.31 Lillian 66.14 Nora 67.54 Zoey 63.45 Mila Answer only for question 5, please

Answer 1

Note: Along with Question 5 , the parts required to understand Question 5 have been solved here.

2.

Mean()= 65, Standard Deviation()= 3.5

a) The height of elizabeth= 65.4

The z score for elizabeth== (65.4- 65)/2.5= 0.16

c) The probability that a randomly selected female is taller than Elizabeth.

P(Z>0.16)= 0.43644

4) Sample mean of 30 female heights= 65.4,

The standard deviation for sampling distribution(SE)= = = 0.64

Z value for 65.4== (65.4- 65)/0.64= 0.626

P(>65.4 )= P(Z>0.626)= 0.2657

5)

The height of elizabeth is equal to the given sample mean height which is 65.4,

So we are calculating probabilities of the heights to be greater than or equal to 65.4.

The only difference is , in case of 2c, we are considering the distribution of individual heights which has standard deviation() of 3.5 ,

But in case of 4, we are considering the sampling distribution instead of individual heights. So in this case the standard deviation has reduced to = 0.64. As the standard deviation has reduced, the distribution will be closer to the mean and Z value will be more than earlier case, so for a sample to have mean height greater than 65.4 will have low probability.

Observe the below standard normal graphs for both cases

Case 2c

Case 4