Question

According to a marketing​ website, adults in a certain country average 54 minutes per day on mobile devices this year. Assume that minutes per day on mobile devices follow the normal distribution and has a standard deviation of 9 minutes. Complete parts a through d below.

a. What is the probability that the amount of time spent today on mobile devices by an adult is less than 63 ​minutes? nothing ​(Round to four decimal places as​ needed.)

b. What is the probability that the amount of time spent today on mobile devices by an adult is more than 49 ​minutes? nothing ​(Round to four decimal places as​ needed.)

c. What is the probability that the amount of time spent today on mobile devices by an adult is between 39 and 52 ​minutes? nothing ​(Round to four decimal places as​ needed.)

d. What amount of time spent today on mobile devices by an adult represents the 75th ​percentile? An amount of time of nothing minutes represents the 75th percentile.

Answer 1

Solution-A:

X~N(54,9)

P(X<63)

Rcode:

library(tigerstats)
pnormGC(bound=63,region="below", mean=54,sd=9,graph=TRUE)

0.8413

Soluiton-b:

P(X>49)

library(tigerstats)
pnormGC(bound=49,region="above", mean=54,sd=9,graph=TRUE)

0.7107

Solution-c:

P(39<X<52)

Rcode;

pnormGC(bound=c(39,52),region="between",mean=54,sd=9,graph=TRUE)

0.3643

Solution-d:

Rocde:

qnormGC(0.75,region="below",mean=54,sd=9,graph=TRUE)

60.07