Chapter 6, Problem 3TP

Solution

To State:

The approximate probability that four randomly selected women are all between $58.6$ inches and $66.7$ inches tall.

Option B $44\%$ is the only correct answer.

Given:

The heights of adult women in the United States are normally distributed with a mean of $64$ inches and a standard deviation of $2.7$ inches.

The Options:

1. $34\%$
2. $44\%$
3. $68\%$
4. $81.5\%$

Concepts Used:

The z-score of the value $x$ is $\frac{x-\overline{x}}{\sigma }$ , where $\overline{x}$ is the mean and $\sigma$ is the standard deviation.

The probability that the z-score of a randomly chosen value from a distribution lies between $\pm 1$ is $68\%$ . The probability that the z-score of a randomly chosen value from a distribution lies between $\pm 2$ is $95\%$ . The probability that the z-score of a randomly chosen value from a distribution lies between $\pm 3$ is $99.7\%$ .

The standard normal curve is symmetrical about the mean so it follows that the probability of a random value to have a z-score between $\pm x$ is double of the probability of the z-score to lie between $0$ and $x$ .

The probability of two independent events happening together is the product of their individual probabilities.

Calculations:

Determine the z-score for $58.6$ inches and $66.7$ inches. These are $\frac{58.6-64}{2.7}=-2$ and $\frac{66.7-64}{2.7}=1$ .

The probability that the z-score of a randomly chosen value from a distribution lies between $-2$ and $1$ is $\frac{95\%}{2}+\frac{68\%}{2}=81.5\%$ .

For a sufficiently large population the range where a particular selection lies is independent of the other selections, thus, the probability that four selected women all lie in the same range is ${\left(81.5\%\right)}^4\approx 44\%$ .

Conclusion:

$44\%$ is the approximate probability that four randomly selected women are all between $58.6$ inches and $66.7$ inches tall . Only Option B satisfies the obtained result so it is the only correct answer.