Question

Make sure that all statistical analysis to be done in Excel and/or StatCrunch and answer all parts:

You will be performing an analysis on heights in the US population, broken out by gender. You will need to know that US heights for males and females both follow an approximately normal distribution. The average height for women is 63.7 inches and a standard deviation of 2.7 inches. The average height for men is 69.1 inches and a standard deviation of 2.9 inches. You will use these numbers in your calculations.

Steps (all statistical analysis to be done in Excel and/or StatCrunch):

1. Collect the heights from 5 females and 5 males that you know. Make sure that you put all heights into inches. Include yourself in the analysis (as one of the 10).

2. Normal Distribution application:

a. Calculate z-scores for each of the selected people and provide interpretations for at least 2 of your z-scores (after looking up probabilities on the table). You also need to include interpretations that indicate how “rare” your observations are in the body of your paper. (Ex: Individual A from my sample with a height of ______ had a corresponding z-score of _____. This means that _______% of people are taller/shorter than him/her)

b. Include one graphic that shows the normal curve along with labels for the middle 68%, 95% and 99.7%. Also show where your own height is on the curve along with an interpretation of how relatively “tall” or “short” you are in comparison with the rest of your gender.

c. Calculate the mean of your male heights and female heights and compare this with the population mean (is it above/below – why is it not exact).

d. Calculate z-scores for the average heights of WNBA and NBA players (typically much taller than the average person) and also male jockeys (typically quite a bit shorter than the average person). Provide interpretations for each.

3. Binomial Experiment application:

a. In order to be considered for a tier 1 point guard in women’s basketball, you need to be 5’8”. Let’s consider a “success” as finding a woman who is 5’8” or taller. The probability of success is approximately 5%. If we consider a sample of 1000 women, calculate the mean and the standard deviation of the binomial random variable.

b. Calculate the range of “normal” observations (not unusual) for the number of women that we would expect to be 5’8” or taller in a sample of 1000 women and use it in a sentence.

c. Use a binomial calculator to calculate the probability of selecting*:

i. Less than 40 women who are 5’8” or taller

ii. Exactly 60 women who are 5’8” or taller

iii. Between 50 and 80, inclusive, women who are 5’8” or taller

iv. More than 70 women who are 5’8” or taller

v. At least 60 women who are 5’8” or taller

*Don’t just give answers here, use complete sentences. Graphs would be a nice addition here (StatCrunch screen grabs or Excel graphs).

d. Can we use the normal distribution as an approximation for the binomial in this case? Why or why not? If yes, what is the probability that we would choose less than 40 women who are 5’8” or taller using this approximation? How does this value compare with the value calculated in part c(i) above?

Answer 1

1) Collected height:

Female	Z score	Male	Z score
62.7 63.5 64.2 61.3 65.4	-0.37037 -0.07407 0.185185 -0.88889 0.62963	71.3 68.5 70.4 67.9 70.8	0.758621 -0.2069 0.448276 -0.41379 0.586207
mean= 63.42		mean=69.78

2) Normal Distribution application:

a) The last woman with a height of 65.4 inches had a corresponding z score of 0.63. From the probability table we can find that she is taller than 73.57% of the female population. and also (0.7357-0.5)=0.2357 i.e. 23.57% of the female are taller than the mean female height but shorter than the last woman.

The first man with a height of 71.3 inches had a corresponding z score of 0.76. From the probability table we can find that he is taller than 77.64% of the male population. and also (0.7764-0.5)=0.2764 i.e. 27.64% of the male are taller than the mean male height but shorter than the first man.

C) for my data The mean female height is 63.42 and mean male height is 69.78. Hence z scores are -0.10 and 0.23 respectively and the corresponding probability from table are 0.4602 and 0.5910. The female mean height is below and mean male height is above than the respective population means. These are not exact as because we are dealing with onle one sample. If we consider all possible samples, then the result will be exact.

D) The average heights of WNBA and NBA players are 72 inches and 79 inces respectively. then for WNBA, zscore is = (72-63.7)/2.7=3.07 and for NBA, zscore is (79-69.1)/2.9=3.41. for WNBA corresponding probability is 0.9989 and for NBA, it is 0.9997. We can conclude that, WNBA players are taller than 99.89% of the female population while NBA players are taller than 99.97% of the male population.

Assuming average height of male jockey is 60 inches. Then zscore is -3.13 and corresponding probability is 0.0009. Therefore we can conclude that less than 1% of male population are shorter than the male jockey with an average height of 60 inches.

3)

a) for binomial distribution with parameter n and p, where n = number of observations and p = probability of success, mean = np and variance = np(1-p). hence mean= 1000*0.05=50 and variance = 1000*.05*0.95=47.5. Hence standard deviation is sqrt(variance)= 6.89

b) woman with a height of 68 inches, the z score is (68-50)/6.89=2.61 and corresponding probability is 0.9955. Hence (1-0.9955)=0.0045 i.e. 0.45% of women have heights 60 inches or more. Hence the number of women haveng height 60 incher or more is 5 approximately.

C)

i) P(X<40)=0.059, i.e. probability of selecting less than 40 women who are 58 inches or taller is 0.59

ii) probability of selecting exactly 60 women who are 5’8” or taller i.e. P(X=60)=0.019

iii) probability of selecting, between 50 and 80, inclusive, women who are 5’8” or taller is P(50<=X<=80) =

P(X<=80) - P(X<=50) = 0.462

iv) probability of selecting, more than 70 women who are 5’8” or taller P(X>70)=0.002.

v) probability of selecting, at least 60 women who are 5’8” or taller P(X>=60)=0.086.

d.) Yes we can use a normal approximation as the total number of observation is 1000 which is large enogh. and the corresponding z score is -1.45 while the probability that we would choose less than 40 women who are 5’8” or taller is 0.0735