Question

A study suggested that childrenbetween the ages of 6 and 11 in the US have anaverage weightof 74 lbs. with a standard deviation of 2.7 lbs.If we assumethe weights are normally distributed, what proportion of childrenin this age range between 70 lbs. and 85 lbs.

Answer 1

Answer:

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.

This means that $\mu = 74, \sigma = 2.7$

What proportion of childrenin this age range between 70 lbs and 85 lbs.

This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So

X = 85

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{85 - 74}{2.7}$

$Z = 4.07$

$Z = 4.07$ has a pvalue of 1

X = 70

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{70 - 74}{2.7}$

$Z = -1.48$

$Z = -1.48$ has a pvalue of 0.0694

1 - 0.0694 = 0.9306

The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.