Respuesta :
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 [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.
For a proportion p in a sample of size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem:
[tex]\mu = 0.4, \sigma = \sqrt{\frac{0.4*0.6}{500}} = 0.0219[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.44 - 0.4}{0.0219}[/tex]
[tex]Z = 1.83[/tex]
[tex]Z = 1.83[/tex] 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%.
