Question

Assume the heights in a female population are normally distributed with mean 65.7 inches and standard deviation 3.2 inches. Then the probability that a typical female from this population is between 5 feet and 5 feet 7 inches tall is (to the nearest three decimals) which of the following?

a. 0.620
b. 0.658
c. 0.963

Answer 1

Choice a. approximately
`0.62`.

Explanation:

This explanation shows how to solve this problem using a typical
`z`-score table. Consider a normal distribution with mean
`\mu` and variance
`\sigma^2`. The
`z\!`-score for a measurement of value
`x` would be
`(x - \mu) / \sigma`.

Convert all heights to inches:

• 
  `5\; \rm ft = 5 * 12 \; in = 60\; \rm in`.
• 
  `5\; \rm ft + 7\; \rm in = 5 * 12 \; in + 7\; \rm in = 67\; \rm in`.

Let
`X` represent the height (in inches) of a female from this population. By the assumptions in this question:
`X \sim \mathrm{N}(65.7,\, 3.2)`. The question is asking for the probability
`P(60 \le X \le 67)`. Calculate the
`z` score for the two boundary values:

• For the lower bound,
  `60\; \rm in`:
  `\displaystyle z = (60 - 65.7)/(3.2) \approx -1.78`.
• For the upper bound,
  `67\; \rm in`:
  `\displaystyle z = (67 - 65.7)/(3.2) \approx 0.41`.

Look up the corresponding probabilities on a typical
`z`-score table.

For the
`z`-score of the upper bound, the corresponding probability is approximately
`0.6591`. In other words:

`P(x \le 67) \approx 0.6591`

On the other hand, some
`z`-score table might not include the probability for negative
`z\!` scores. That missing part can be found using the symmetry of the normal distribution PDF.

The probability corresponding to
`P(z < 1.78)` (that's the opposite of the
`z\!`-score at the lower bound) is approximately
`0.9625`. By the symmetry of the normal PDF:

`P(z < -1.78) = 1 - P(z < 0 - (-1.78)) \approx 1 - 0.9625 = 0.0375`.

Therefore:

`P(X < 60) \approx 0.0375`.

Calculate the probability of the interval between the two bounds:

`\begin{aligned}P(60 \le X \le 65.7) &= P(X \le 65.7) - P(X \le 60)\\ &\approx 0.6591 - 0.0375 \approx 0.62 \end{aligned}`.