Answer:
The probability that 44% or fewer of the pre-teens have a TV in their bedroom is 96.64%.
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.
For a proportion p in a sample of size n, we have that
In this problem:
The probability that 44% or fewer of the pre-teens have a TV in their bedroom is
This is the pvalue of Z when X = 0.44. So
has a pvalue of 0.9664
The probability that 44% or fewer of the pre-teens have a TV in their bedroom is 96.64%.
