Answer:
1) what percent of men are between 64.5 and 73.5 inches tall?
= 95.45%
2) what percent of men are taller than 71.25 inches?
= 15.866%
3) what is your z score? for ladies N(64,2)
= 2.5
4) what is the height that 90% of men in the us are taller than?
66.12 inches
Explanation:
We solve using the z score formula
The formula for calculating a z-score is is z = (x-μ)/σ,
where x is the raw score
μ is the population mean
σ is the population standard deviation.
a) what percent of men are between 64.5 and 73.5 inches tall?
mean 69 inches and standard deviation 2.25.
For x = 64.5
z = (x-μ)/σ
= 64.5 - 69/2.25
= -2
Probability value from Z-Table:
P(x= 64.5) = 0.02275
For x = 73.5
z = (x-μ)/σ
= 73.5 - 69/2.25
= 2
Probability value from Z-Table:
P(x = 73.5) = 0.97725
Hence, percent of men are between 64.5 and 73.5 inches tall
= 64.5<x < 74.5
P(x = 73.5) - P(x = 64.5)
= 0.97725 - 0.02275
= 0.9545
Converting to percentage = 0.9545
× 100
= 95.45%
b) what percent of men are taller than 71.25 inches?
mean 69 inches and standard deviation 2.25.
For x = 64.5
z = (x-μ)/σ
= 71.25 - 69/2.25
= 1
Probability value from Z-Table:
P(x<71.25) = 0.84134
P(x>71.25) = 1 - P(x<71.25)
= 0.15866
Converting to percentage
= 0.15866 × 100
= 15.866%
The percent of men are taller than 71.25 inches is 15.866%
3) what is your z score? for ladies N(64,2)
z = (x-μ)/σ
Mean = 64, standard deviation = 2
x = 69
z = 69 - 64/2
z = 5/2
z = 2.5
4) what is the height that 90% of men in the us are taller than?
10th percentile confidence interval = 90% of men = -1.282
z = (x-μ)/σ
-1.282 = x - 69/2.25
-1.282 × 2.25 = x - 69
2.8845 = x - 69
x = 69 - 2.8845
x = 66.1155
Approximately = 66.12 inches