Respuesta :
Answer:
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 [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.
In this problem, we have that:
[tex]\mu = 60, \sigma = 25[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 60}{25}[/tex]
[tex]Z = 2.4[/tex]
[tex]Z = 2.4[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 60}{25}[/tex]
[tex]Z = -1.2[/tex]
[tex]Z = -1.2[/tex] has a pvalue of 0.1151.
11.51% of 2-year-old children watch less than 30 minutes a day.
