Question

According to a report by App Annie, a business intelligence company that produces tools and reports for the apps and digital goods industry, smartphone owners are using an average of 30 apps per month.Assume that number of apps used per month by smartphone users is normally distributed and that the standard deviation is 5. If you select a random sample of 25 smartphone owners,a. what is the probability that the sample mean is between 29 and 31?b. what is the probability that the sample mean is between 28 and 32?c. If you select a random sample of 100 smartphone owners, what is the probability that the sample mean is between 29 and 31?d. Explain the difference in the results of (a) and (c).

Answer 1

Answer:

a) $P(29\leq \bar X >31)=P(\frac{29-30}{1}\leq Z\leq \frac{31-30}{1})=P(-1\leq Z \leq 1)=P(Z<1)-P(Z<-1) =0.841-0.159=0.683$

b) $P(28\leq \bar X >32)=P(\frac{28-30}{1}\leq Z\leq \frac{32-30}{1})=P(-2\leq Z \leq 2)=P(Z<2)-P(Z<-2) =0.977-0.0228=0.955$

c) $P(29\leq \bar X >31)=P(\frac{29-30}{0.5}\leq Z\leq \frac{31-30}{0.5})=P(-2\leq Z \leq 2)=P(Z<2)-P(Z<-2) =0.977-0.0228=0.95$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable number of apps used per month by smartphone users. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =5)$

We take a sample of n=25 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{5}{\sqrt{25}})$

a. what is the probability that the sample mean is between 29 and 31?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(29\leq \bar X >31)=P(\frac{29-30}{1}\leq Z\leq \frac{31-30}{1})=P(-1\leq Z \leq 1)=P(Z<1)-P(Z<-1) =0.841-0.159=0.683$

b. what is the probability that the sample mean is between 28 and 32?

$P(28\leq \bar X >32)=P(\frac{28-30}{1}\leq Z\leq \frac{32-30}{1})=P(-2\leq Z \leq 2)=P(Z<2)-P(Z<-2) =0.977-0.0228=0.955$

c. If you select a random sample of 100 smartphone owners, what is the probability that the sample mean is between 29 and 31?

On this case we have a new distribution for the sample mean given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{5}{\sqrt{100}})$

And we want to find this probability:

$P(29\leq \bar X >31)=P(\frac{29-30}{0.5}\leq Z\leq \frac{31-30}{0.5})=P(-2\leq Z \leq 2)=P(Z<2)-P(Z<-2) =0.977-0.0228=0.95$

On this case the results for parts a and b shows the probability that we have values within one and two deviations from the mean. And the result for part c its equal for part b since the new deviation for the sampel mean with a sample size of 100 its the half of the standard deviation when we use a random sample of 25.