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

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

Convert all heights to inches:

[tex]5\; \rm ft = 5 \times 12 \; in = 60\; \rm in[/tex].
[tex]5\; \rm ft + 7\; \rm in = 5 \times 12 \; in + 7\; \rm in = 67\; \rm in[/tex].

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

For the lower bound, [tex]60\; \rm in[/tex]: [tex]\displaystyle z = \frac{60 - 65.7}{3.2} \approx -1.78[/tex].
For the upper bound, [tex]67\; \rm in[/tex]: [tex]\displaystyle z = \frac{67 - 65.7}{3.2} \approx 0.41[/tex].

Look up the corresponding probabilities on a typical [tex]z[/tex]-score table.

For the [tex]z[/tex]-score of the upper bound, the corresponding probability is approximately [tex]0.6591[/tex]. In other words:

[tex]P(x \le 67) \approx 0.6591[/tex]

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

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

[tex]P(z < -1.78) = 1 - P(z < 0 - (-1.78)) \approx 1 - 0.9625 = 0.0375[/tex].

Therefore:

[tex]P(X < 60) \approx 0.0375[/tex].

Calculate the probability of the interval between the two bounds:

[tex]\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}[/tex].