A study indicates that 18- to 24- year olds spend a mean of 130 minutes watching video on their smartphones per month. Assume that the amount of time watching video on a smartphone per month is normally distributed and that the standard deviation is 20 minutes. Complete parts (a) through (d) below.

Question content area bottom

Part 1

a.

What is the probability that an 18- to 24-year-old spends less than 100 minutes watching video on his or her smartphone per month? (Round to four decimal places as needed.)

A.

P(X<100)=NORM.DIST(99,130,20,1)=0.6057

B.

P(X<100)=NORM.DIST(130,100,20,1)=0.9332

C.

P(X<100)=NORM.DIST(100,130,20,1)=0.0668

D.

P(X<100)=NORM.DIST(130,130,20,1)=0.5000

b.

What is the probability that an 18- to 24-year-old spends between 100 and 140 minutes watching video on his or her smartphone per month? (Round to four decimal places as needed.)

A.

P(100

B.

P(100

C.

P(100

D.

P(100

-0.6247

c.

What is the probability that an 18- to 24-year-old spends more than 140 minutes watching video on his or her smartphone per month? (Round to four decimal places as needed.)

A.

P(X>140)=NORM.DIST(140,130,20,1)=0.6915

B.

P(X>140)=1-NORM.DIST(140,130,20,0)=0.9824

C.

P(X>140)=1-NORM.DIST(140,130,20,1)=0.3085

D.

P(X>140)=1-NORM.DIST(141,130,20,1)=0.2912

d.

Five percent of all 18- to 24-year-olds will spend less than how many minutes watching video on his or her smartphone per month? (Round to two decimal places as needed.)

A.

P(X)=0.05 --> 100-NORM.INV(0.05,130,20)=2.90

B.

P(X)=0.05 --> NORM.DIST(0.05,130,20)=0.00

C.

P(X)=0.05 --> 1.00 - 0.05 = 0.95 --> NORM.INV(0.95,130,20)=162.90

D.

P(X)=0.05 --> NORM.INV(0.05,130,20)=97.10