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Zombio · Answer 1 · 2021-11-18T00:09:45+0000

Answer:

Choice a. approximately
0.62 .

Explanation:

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

For the
-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
-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:

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

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