Question

Use Z Table only (not Excel)

The Bank of America Trends in Consumer Mobility Report indicates that in a typical day, 51% of users of mobile phones use their phone at least once per hour, 26% use their phone a few times per day, 8% use their phone morning and evening, and 13% hardly ever use their phones. The remaining 2% indicated that they did not know how often they used their mobile phone (USA Today, July 7, 2014). Consider a sample of 150 mobile phone users.

a)What is the probability that at least 70 use their phone at least once per hour?

b)What is the probability that at least 75 but less than 80 use their phone at least once per hour?

c)What is the probability that less than 5 of the 150 phone users do not know how often they use their phone?

Answer 1

a)

n=

150

p=

0.5100

here mean of distribution=μ=np=		76.50
and standard deviation σ=sqrt(np(1-p))=		6.12
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

probability that at least 70 use their phone at least once per hour :

probability =P(X>69.5)=P(Z>(69.5-76.5)/6.122)=P(Z>-1.14)=1-P(Z<-1.14)=1-0.1271=0.8729

b)

probability =P(74.5<X<79.5)=P((74.5-76.5)/6.122)<Z<(79.5-76.5)/6.122)=P(-0.33<Z<0.49)=0.6879-0.3707=0.3172

c)

n=

150

p=

0.0200

here mean of distribution=μ=np=		3.00
and standard deviation σ=sqrt(np(1-p))=		1.71

probability =P(X<4.5)=(Z<(4.5-3)/1.715)=P(Z<0.87)=0.8078