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

Answer:

Choice a. approximately $0.62$.

Step-by-step 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:

• $5\; \rm ft = 5 \times 12 \; in = 60\; \rm in$.
• $5\; \rm ft + 7\; \rm in = 5 \times 12 \; in + 7\; \rm in = 67\; \rm in$.

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, $60\; \rm in$: $\displaystyle z = \frac{60 - 65.7}{3.2} \approx -1.78$.
• For the upper bound, $67\; \rm in$: $\displaystyle z = \frac{67 - 65.7}{3.2} \approx 0.41$.

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:

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