A branch of the military requires women's heights to be between 58 in and 80 in. We need to find the probabilities that heights fall between 58 in and 80 in in this distribution. We need to find z-scores of the values 58 in and 80 in. Z-score shows how many standard deviations far are the values from the mean. Therefore they subtracted from the mean and divided by the standard deviation:

z-score of 58 in= $z=\frac{58-63.9}{2.4}$ = -2.458

z-score of 80 in= $z=\frac{80-63.9}{2.4}$ = 6.708

In normal distribution 99.3% of the values have higher z-score than -2.458

0% of the values have higher z-score than 6.708. Therefore 99.3% of the woman meet the height requirement.

To find the height requirement so that all women are eligible except the shortest 1% and the tallest 2%, we need to find the boundary z-score of the

shortest 1% and the tallest 2%. Thus, upper bound for z-score has to be 2.054 and lower bound is -2.326

Corresponding heights (H) can be found using the formula

$2.054=\frac{H-63.9}{2.4}$ and

$-2.326=\frac{H-63.9}{2.4}$

Thus lower bound for height is 58.32 and

Upper bound for height is 68.83

8 0

3 years ago

If we wanted to explore the idea that mother's smoking might explain the variation in birth weight, what visualization might be

Mila [183]

Answer:

If a mother is smoking and there are significant birth weights, a bar graph will be most effective. A bar graph can show how many children have similar weights, and you can tell which weights are most common.

Hope this helped.

5 0

3 years ago