Question

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 36% and justify your answer.

b) Find the 75th percentile, and express it's meaning in a complete sentence.

Answer 1

p=0.28

SD = 0.05

a.

b.

75th percentile:

p(z<Z)

Z = 0.67 {from table}

p = 0.28+0.05*0.67 = 0.3135

this means 75% chance that less than 31.35% of 18 to 34 year olds check their Facebook profiles before getting out of bed in the morning

(please UPVOTE)

P(z<Z) table :