im begging pls help ive gotten this wrong too many times

Question

The heights of adult men in America are normally distributed, with a mean of 69.2 inches and a standard deviation of 2.69 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.2 inches and a standard deviation of 2.54 inches.

a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?

z = 

b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.

%

c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

z = 

d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.

%

e) Who is relatively taller: a 6'3" American man or a 5'11" American woman? Defend your choice in a meaningful sentence.

Answer 1

A) If a man is 6 feet 3 inches tall, his z-score will be:

z = (x-μ)/σ = (75-69.2)/2.69 ≈ 2.16

B) The percentage of men that are SHORTER than 6 feet 3 inches can be represented with:

P(X<75) = P(Z<2.16) ≈ 0.9846 ≈ 98.5%

C) If a woman is 5 feet 11 inches tall, her z-score will be:

z = (x-μ)/σ = (71-64.2)/2.54 ≈ 2.68

D) The percentage of women that are TALLER than 5 feet 11 inches can be represented with:

P(Z>2.68) ≈ 1 - 0.9963 = 0.0037 = 0.37%

E) Long answer: To figure out who is relatively taller, we would need to compare the percentages of the American man and woman's heights relative to the normal/standard distribution of both gender's heights. The percentage of men SHORTER than 6'3" is about 98.5%, and given that 0.37% of women are TALLER than 5'11", then 100%-0.37% = 99.63% of women are SHORTER than 5'11". Because 99.63%>98.5%, then a 5'11" American woman would be relatively taller.

Shorter answer: There is a greater percentage of women shorter than 5'11" than there are of men who are shorter than 6'3". For women, this percentage is 100%-0.37%=99.63%, and for men, this number is 98.5%.

Note that all z-scores were determined by a z-table, so the answers may be off by a very small decimal/percentage amount.

Hope this helps!

Answer 2

z-score is a measure of how many standard deviations something is from the mean. Yo can use the equation:

where "z" is the z-score, "x" is the observed value, μ is the mean, and σ is the standard deviation

a) If a man is 6' 3", that means his height in inches is 6(12) + 3 = 75 inches so plugging in x = 75, μ = 69.2 and σ = 2.69, we get:

(75 - 69.2)/2.69 = 5.8/2.69 = 2.156133829 which rounds to 2.16

b) To answer this question using a TI-84 Plus CE calculator, use the "distr" button (which is the blue "2nd" button then "vars" which is 3 buttons up from the 9). After you hit "distr", scroll to "2:normalcdf(" then hit enter. You will get a screen asking you for "lower", "upper", "μ", and "σ" and will say "Paste" on the bottom. For "lower" type in -1000000

For "upper" type 75

For "μ" type 69.2

For "σ" type 2.69

hit "enter on "Paste" then hit "enter a second time.

You should get 0.9844634475

Convert this into a % by multiplying by 100. So the answer is 98.4% of men are shorter than this.

c) Again, convert to inches: 5' 11" = 5(12) + 11 = 71 inches then use the z-score equation z = (71 - 64.2)/2.54 = 6.8/2.54 = 2.68

d) Again, to answer this question using a TI-84 Plus CE calculator, use the "distr" button. After you hit "distr", scroll to "2:normalcdf(" then hit enter. This time (since you want the % TALLER)

For "lower" type in 71

For "upper" type 1000000

For "μ" type 64.2

For "σ" type 2.54

hit "enter on "Paste" then hit "enter a second time.

You should get 0.0037124452

Convert this into a % by multiplying by 100. So the answer is 0.4% of women are taller than this.

e) You can answer this question by comparing the z-scores. A man that is 75 inches has a z-score of 2.16 while a woman that is 71 inches has a z-score of 2.68 meaning the woman is farther away from the mean than the man so the woman is "relatively" taller