Question

A consumer mobility report indicates that in a typical day, 51% of users of mobile phones use their phone at least once per hour, 25% use their phone a few times per day, 8% use their phone morning and evening, and 12% hardly ever use their phones. The remaining 4% indicated that they did not know how often they used their mobile phone. Consider a sample of 180 mobile phone users.

(a)

What is the probability that at least 90 use their phone at least once per hour? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

(b)

What is the probability that at least 95 but less than 100 use their phone at least once per hour? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

(c)

What is the probability that less than 10 of the 180 phone users do not know how often they use their phone? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)

Answer 1

a)
mean = 0.51*180 = 91.8
sd = sqrt(180*0.51*0.49) = 6.7069

P(X >= 90)
= P(X > 89.5) .. continuity correction applied
= P(z > (89.5 - 91.8)/6.7069)
= P(z > -0.3429)
= 0.6342

b)
P(95 <= X < 100)
= P(94.5 < X < 99.5) .. continuity correction
= P((94.5 - 91.8)/6.7069 < z < (100 - 91.8)/6.7069))
= P(0.4026 < z < 1.2226)
= 0.2329

c)
mean = 0.04*180 = 7.2
sd = sqrt(180 * 0.04 * 0.96) = 2.6291

P(X < 10)
= P(X < 9.5)
= P(z < (9.5 - 7.2)/2.6291)
= P(z < 0.8748)
= 0.8092