Question

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

Answer 1

1)

a)

$$\mu_{\bar{X}}=\mu=8.2$$

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

b)

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

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

$$\approx P(Z<2.3238)-P(Z\le-1.5492)$$

$$\approx0.98993-0.06067\approx0.9293$$

2)

a)

$$\mu_{\bar{X}}=\mu=34$$

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

b)

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

$$\approx1-P(Z\le-2.66667)$$

$$\approx0.99617$$