Question

According to a marketing website, adults in a certain country average 55 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 6 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 58 minutes?

B.) What is the probability that the amount of time spent today on mobile devices by an adult is more than 49 minutes?

C.) What is the probability that the amount of time spent today on mobile devices by an adult is between 39 and 54 minutes?

D.) What amount of time spent today on mobile devices by an adult represents the 85th percentile?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	55
std deviation   =σ=	6

probability that the amount of time spent today on mobile devices by an adult is less than 58 minutes:

probability =

P(X<58)

=

P(Z<0.5)=

0.6915

b)

  probability that the amount of time spent today on mobile devices by an adult is more than 49 minutes:

probability =

P(X>49)

=

P(Z>-1)=

1-P(Z<-1)=

1-0.1587=

0.8413

c)

probability =

P(39<X<54)

=

P(-2.67<Z<-0.17)=

0.4325-0.0038=

0.4287

d)

for 85th percentile critical value of z=		1.04
therefore corresponding value=mean+z*std deviation=			61.24 minute