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:

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu = 74, \sigma = 2.7[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{85 - 74}{2.7}[/tex]

[tex]Z = 4.07[/tex]

[tex]Z = 4.07[/tex] has a pvalue of 1

X = 70

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{70 - 74}{2.7}[/tex]

[tex]Z = -1.48[/tex]

[tex]Z = -1.48[/tex] 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.

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