The probability that the percent of 18 to 34-year-olds who check social media before getting out of bed in the morning is, at most, 32 is 0.7881 = 78.81%
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
In this question:
[tex]\mu=28,\sigma=5[/tex]
Find the probability that the percent of 18 to 34-year-olds who check social media before getting out of bed in the morning is, at most, 32.
This is the p-value of Z when X = 32. So
[tex]Z=\frac{X-\mu}{\sigma}[/tex]
[tex]Z=\frac{32-28}{5}[/tex]
[tex]Z=0.8[/tex]
Z=0.8 has a p-value of 0.7881
0.7881 = 78.81% probability that the percent of 18 to 34-year-olds who check social media before getting out of bed in the morning is, at most, 32.
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