A recent report indicates that 2-year-old children from well-educated suburban families watched an average of μ = 60 minutes of television each day. Assuming that the distribution of television-watching times is normal with a standard deviation of σ = 25 minutes, find each of the following proportions.

a. What proportion of 2-year-old children watch more than 2 hours of television each day?
b. What proportion of 2-year-old children watch less than 30 minutes a day?

a) 0.82% of 2-year-old children watch more than 2 hours of television each day.

b) 11.51% of 2-year-old children watch less than 30 minutes a day.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

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.

In this problem, we have that:

a. What proportion of 2-year-old children watch more than 2 hours of television each day?

This is 1 subtracted by the pvalue of Z when X = 2*60 = 120 minutes.

So

has a pvalue of 0.9918

1 - 0.9918 = 0.0082

0.82% of 2-year-old children watch more than 2 hours of television each day.

b. What proportion of 2-year-old children watch less than 30 minutes a day?

This is the pvalue of Z when X = 30. So

has a pvalue of 0.1151.

11.51% of 2-year-old children watch less than 30 minutes a day.

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.