Question

In: Statistics and Probability

According to a social media​ blog, time spent on a certain social networking website has a...

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.

Solutions

Expert Solution

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.


Related Solutions

According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 21 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 3 minutes. Complete parts​ (a) through​ (d) below. a. If you select a random sample of 16 ​sessions, what is the probability that the sample mean is between 20.5 and 21.5 ​minutes? nothing ​(Round to three decimal places...
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 22 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes. Complete parts​ (a) through​ (d) below. a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 21.5 and 22.5 minutes? (Round to three decimal places) b.If...
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 18 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 17.5 and 18.5 ​minutes?
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 18 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 4 minutes. a. If you select random sample of 25 sessions, what is the probability that the sample mean is between 17 and 18 minutes?
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 17 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 7 minutes. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 16.5 and 17.5 minutes? A global research study found that the majority of​ today's working women would...
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 19 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 44 minutes. Complete parts​ (a) through​ (d) below a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 18.5 and 19.5 ​minutes? b. If you select a random...
According to a social media​ blog, time spent on a certain social networking website has a...
According to a social media​ blog, time spent on a certain social networking website has a mean of 22 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 7 minutes. Complete parts​ (a) through​ (d) below. a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 21.5 and 22.5 ​minutes? ___ ​(Round to three decimal places...
2. A researcher is interested in how time spent on social media is related to self-esteem....
2. A researcher is interested in how time spent on social media is related to self-esteem. He has participants how much time they spend on social media per day and has them take a self-esteem scale (on an 11 point scale). The data is as follows: participants hours per day on social media self-esteem score A 7 4 B 4 7 C 8 9 D 3 10 E 6 8 F 2 11 G 3 10 H 5 5 b....
Technology Research Blog Social Media has increasingly become an important tool for businesses to connect with...
Technology Research Blog Social Media has increasingly become an important tool for businesses to connect with their stakeholders. A blog (short form of "weblog") is a social media tool that helps in engaging in discussion with the stakeholders through information published on the World Wide Web. The blogs are normally used to develop customer relationships by driving traffic to the business website and increasing the business’s search engine optimization (SEO). Blogging is integral to the online marketing strategy of all...
What impact has modern technology, such as the Internet, media, social networking, advertising, had on eating...
What impact has modern technology, such as the Internet, media, social networking, advertising, had on eating disorders in children and adolescents? How can such be used in a positive way rather than causing harm?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT