Question #343114

The mean number of minutes that android phone users spend on an app is 8.2 minutes. Suppose the standard deviation is 1 minute. Take a sample of 60 What are the mean, standard deviation for the sample mean number of app engagement by a tablet user? and Find the probability that the sample mean is between 8 and 8.5 minutes. 2.) In a recent study, the mean age of online job search website users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size n=100. a.) What are the mean, and standard deviation for the sample mean ages of tablet users? Find the probability that the sample mean age is more than 30 years


1
Expert's answer
2022-05-22T17:52:03-0400

1)

a)


μXˉ=μ=8.2\mu_{\bar{X}}=\mu=8.2

σXˉ=σ/n=1/600.1291\sigma_{\bar{X}}=\sigma/\sqrt{n}=1/\sqrt{60}\approx0.1291

b)


P(8<Xˉ<8.5)=P(Z<8.58.21/60)P(8<\bar{X}<8.5)=P(Z<\dfrac{8.5-8.2}{1/\sqrt{60}})

P(Z88.21/60)-P(Z\le\dfrac{8-8.2}{1/\sqrt{60}})

P(Z<2.3238)P(Z1.5492)\approx P(Z<2.3238)-P(Z\le-1.5492)

0.989930.060670.9293\approx0.98993-0.06067\approx0.9293

2)

a)


μXˉ=μ=34\mu_{\bar{X}}=\mu=34

σXˉ=σ/n=15/100=1.5\sigma_{\bar{X}}=\sigma/\sqrt{n}=15/\sqrt{100}=1.5

b)


P(Xˉ>30)=1P(Z303415/100)P(\bar{X}>30)=1-P(Z\le\dfrac{30-34}{15/\sqrt{100}})

1P(Z2.66667)\approx1-P(Z\le-2.66667)

0.99617\approx0.99617


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS