Facebook provides a variety of statistics on its Web site that detail the growth and popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
b. Find the 95th percentile, and express it in a sentence.
a. P(X >=0.30), calculate the z-score which is:
Z = 0.4
P(x>0.4) = 0.344578 or 34.46%
b. Inverse Normal (0.95) calculator = 1.644853627
Use NORMSINV(0.95) on Excel
0.28 + 0.05(1.644853627) = 0.362242681 or 36.22%
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before getting out of bed in the morning is at least 30.
b. Find the 95th percentile, and express it in a sentence.
a. P(X >=0.30), calculate the z-score which is:
Z = 0.4
P(x>0.4) = 0.344578 or 34.46%
b. Inverse Normal (0.95) calculator = 1.644853627
Use NORMSINV(0.95) on Excel
0.28 + 0.05(1.644853627) = 0.362242681 or 36.22%