Question

According to a social media​ blog, time spent on a certain social networking website has a mean of 16 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 5 minutes. Complete parts​ (a) through​ (d) below.

a. If you select a random sample 36 sessions, what is the probability that the sample mean is between 15.5 and 16.5 minutes? = 0.452

b. If you select a random sample of 36 sessions, what is the probability that the sample mean is between 15 and 16 minutes?= 0.385

c. If you select a random sample of 144 ​sessions, what is the probability that the sample mean is between 15.5 and 16.5 ​minutes?= 0.769

d. Explain the difference in the results of​ (a) and​ (c).

The sample size in​ (c) is greater than the sample size in​ (a), so the standard error of the mean​ (or the standard deviation of the sampling​ distribution) in​ (c) is (greater/less)-(which one) than in​ (a). As the standard error (decreases/increases)- (which one) values become more concentrated around the mean.​ Therefore, the probability that the sample mean will fall in a region that includes the population mean will always (decrease/increase)- (which one) when the sample size increases.

Answer 1

Solution:

We are given that the random variable time spent on the social networking site per visit is normally distributed.

Mean = µ = 16

SD = σ = 5

a. If you select a random sample 36 sessions, what is the probability that the sample mean is between 15.5 and 16.5 minutes?

We are given n = 36, µ = 16, σ = 5

We have to find P(15.5<Xbar<16.5)

P(15.5<Xbar<16.5) = P(Xbar<16.5) – P(Xbar<15.5)

First find P(Xbar<16.5)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (16.5 – 16)/[5/sqrt(36)]

Z = 0.5/(5/6)

Z = 0.5/ 0.833333

Z = 0.6

P(Z< 0.6) = P(Xbar<16.5) = 0.725747

(by using z-table)

Now, find P(Xbar<15.5)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (15.5 – 16)/[5/sqrt(36)]

Z = -0.5/(5/6)

Z = -0.5/ 0.833333

Z = -0.6

P(Z<-0.6) = P(Xbar<15.5) = 0.274253

(by using z-table)

P(15.5<Xbar<16.5) = P(Xbar<16.5) – P(Xbar<15.5)

P(15.5<Xbar<16.5) = 0.725747 – 0.274253

P(15.5<Xbar<16.5) = 0.451494

Required probability = 0.451

b. If you select a random sample of 36 sessions, what is the probability that the sample mean is between 15 and 16 minutes?

We are given n = 36, µ = 16, σ = 5

We have to find P(15<Xbar<16)

P(15<Xbar<16) = P(Xbar<16) – P(Xbar<15)

First find P(Xbar<16)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (16 – 16)/[5/sqrt(36)]

Z = 0/(5/6)

Z = 0

P(Z< 0) = P(Xbar<16) = 0.5000

(by using z-table)

Now, find P(Xbar<15)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (15 – 16)/[5/sqrt(36)]

Z = -1/(5/6)

Z = -1/ 0.833333

Z = -1.2

P(Z<-1.2) = P(Xbar<15) = 0.11507

(by using z-table)

P(15<Xbar<16) = P(Xbar<16) – P(Xbar<15)

P(15<Xbar<16) = 0.5000 – 0.11507

P(15<Xbar<16) = 0.38493

Required probability = 0.385

c. If you select a random sample of 144 ​sessions, what is the probability that the sample mean is between 15.5 and 16.5 ​minutes?

We are given n = 144, µ = 16, σ = 5

We have to find P(15.5<Xbar<16.5)

P(15.5<Xbar<16.5) = P(Xbar<16.5) – P(Xbar<15.5)

First find P(Xbar<16.5)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (16.5 – 16)/[5/sqrt(144)]

Z = 0.5/(5/12)

Z = 0.5/ 0.416667

Z = 1.2

P(Z< 1.2) = P(Xbar<16.5) = 0.88493

(by using z-table)

Now, find P(Xbar<15.5)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (15.5 – 16)/[5/sqrt(144)]

Z = -0.5/(5/12)

Z = -0.5/ 0.416667

Z = -1.1999999

P(Z<-1.199999) = P(Xbar<15.5) = 0.11507

(by using z-table)

P(15.5<Xbar<16.5) = P(Xbar<16.5) – P(Xbar<15.5)

P(15.5<Xbar<16.5) = 0.88493 – 0.11507

P(15.5<Xbar<16.5) = 0.769861

Required probability = 0.769

d. Explain the difference in the results of​ (a) and​ (c).

The sample size in​ (c) is greater than the sample size in​ (a), so the standard error of the mean​ (or the standard deviation of the sampling​ distribution) in​ (c) is less than in (a). As the standard error increases, values become more concentrated around the mean.​ Therefore, the probability that the sample mean will fall in a region that includes the population mean will always increase when the sample size increases.