Assume the average height for American women is 64 inches with a standard deviation of 2 inches.

Question

What percent of women is taller than 66 inches?

What percent of American women is less than 68 inches?

What percent of American women is between 60 and 66 inches?

Jenny is taller than 99.8% of American women, what’s jenny’s height in inches?

Answer 1

Let Φ(z)=P(Z<z) represent the cumulative distribution function (CDF) for a standard normal curve.

Percentage of American women taller than 66 in: 1-Φ((66-64)/2) = 1-Φ(1) = 1-0.8413447 = 0.1586553 ≈ 15.87%

Percentage of American women shorter than 68 in: Φ((68-64)/2) = Φ(2) = 0.9772499 ≈ 97.72%

Percentage of American women between 60 and 66 in tall: Φ((66-64)/2)-Φ((60-64)/2) = Φ(1)-Φ(-2) = 0.8185946 ≈ 81.86%

Let z(α) represent the 100(1-α)^th percentile of the standard normal curve. Precisely: 1-α = Φ(z(α)) = Φ((x(α)-μ)/σ) where x(α) is the 100(1-α)^th percentile of the normal curve. Also, let Φ^-1(p)=z be the inverse CDF of the standard normal curve which maps a probability 'p' to its corresponding coordinate 'z' on the standard normal curve.

Let x(0.002) denote Jenny's height, since she is taller than 100(1-0.002)% of American women. From above, we have:

0.998 = Φ((x(0.002)-64)/2); take the inverse CDF of both sides

Φ^-1(0.998) = (x(0.002)-64)/2; now solve for x(0.002) algebraically. Φ^-1(0.002) can be found in a table or calculated using software.

I get x(0.002) = 2(2.878162)+64 = 69.75632. So Jenny is about 69.76 in. tall.

I hope that helps! Let me know if anything is unclear.

Answer 2

For this problem, you need to understand the 68-95-99.7 rule. This states that in a normal distribution, 68% of the data is within 1 standard deviation of the mean, 95% is within 2 standard deviations, and 99.7% is within 3 standard deviations. I suggest you look up a picture of this to visualize what this looks like on a curve.

a) 66 inches is 1 standard deviation away from the mean, so use the 68% part of the rule. To find the percentile for a height of 66 inches, divide 68% by 2 because you are looking at heights greater than the mean, so you get 34%, and then add this to 50% (the mean). This means the percentile is 84. To find the percent of women taller than 66 inches, you subtract 84 from 100 and get 16. This means 16% of American women are taller than 66 inches.

b) 68 inches is 2 standard deviations from the mean, so use the 95% part of the rule. To find the percentile, divide 95% by 2 and add to 50%, so you get 97.5 as the percentile. This means 97.5% of American women are less than 68 inches tall.

c) 60 is 2 standard deviations from the mean and 66 is 1 standard deviation from the mean. 34% of the data (68%/2) falls between 64 and 66, and 47.5% (95%/2) of the data falls between 60 and 64, so add these values to find that 81.5% of American women fall between 60 and 66 inches.

d) The 99.7% rule is for values that are 3 standard deviations from the mean. This means if Jenny is taller than 99.8% of American women, then she is 3 standard deviations above the mean, so her height is 70 inches.