Answer to Question #217762 in Statistics and Probability for PRAGNA

Question

The amount of time a bank teller spends with each customer has a population mean of 3.10 minutes and a standard deviation of 0.40 minutes. Suppose a random sample of 16 customers is selected.

a. What is the distribution of the mean amount of time for the samples?

b. What is the probability that the average time spent per customer will be at least 3 minutes?

c. There is an 85% chance that the sample mean will be below how many minutes?

d. If a random sample of 64 customers is selected, there is an 85% chance that the sample mean will be below how many minutes?

Answer 1

(a). The distribution of the mean amount of samples is shown below;

The normal distribution is a continuous distribution of data that has a bell shaped curve as shown on the above figure. The normally distributed random variable $x$ has the mean of $\mu$ and standard deviation of $\sigma$ . The standardized $z-score$ represents the the number of standard deviations the datapoint is away from the mean.

The $z-score$ takes the positive value when it is above the mean and the negative value when it is below the mean.

(b). $\overline{X}$ denotes the mean time spent for customers which follows a normal distribution with mean 3.10 minutes and the standard deviation of o.40 minutes. To put it into context, $\mu=3.10$ and $\sigma=0.40$ with the sample of 16 customers.

The probability that the average time spent per customer will be at least 3 minutes is:

$\:P\left(\overline{X}\ge 3\right)=1-P\left(\overline{X}<3\right)$

$=1-P\left[\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{3-3.10}{\frac{0.40}{\sqrt{16}}}\right]$

$=1-P\left[z<\frac{-0.1}{0.10}\right]$

$\:=1-P\left(z<-1\right)$

From the standard normal table, the area to the left of $z<-1$ is $0.1587$ .

$P\left(\overline{X}\ge 3\right)=1-P\left(z<-1\right)$

$=1-0.1587$

$=0.8413$

Thus the probability that the mean time spent per customer is at least 3 minutes is 0.8413

(c). The number of minutes for the sample mean with less than the 85% chance is obtained below:

Where $n=16$ and $\sigma \:_{\overline{x}}=0.4$

$P\left(\overline{X}\le \overline{x}\right)=0.85$

$\:P\left[\frac{\overline{X}-\mu _{\overline{x}}}{\frac{\sigma _{\overline{x}}}{\sqrt{n}}}\le \frac{\overline{x}-3.1}{\frac{0.4}{\sqrt{16}}}\right]=0.85$

$P\left(z\le \frac{\overline{x}-3.1}{0.1}\right)=0.85$

The z - value is obtained from the z - table as $z=1.04$

$1.04=\frac{\overline{x}-3.1}{0.1}$

$1.04\times 0.1=\overline{x}-3.1$

$0.104=\overline{x}-3.1$

$\overline{x}=3.1+0.104$

$\overline{x}=3.204$

Thus, the number of minutes required is 3.204.

(d). The number of minutes for the sample mean with less than the 85% chance is obtained below:

Where $n=64$ and $\sigma \:_{\overline{x}}=0.4$ .

$P\left(\overline{X}\le \overline{x}\right)=0.85$

$\:P\left[\frac{\overline{X}-\mu \:_{\overline{x}}}{\frac{\sigma \:_{\overline{x}}}{\sqrt{n}}}\le \:\frac{\overline{x}-3.1}{\frac{0.4}{\sqrt{64}}}\right]=0.85$

$P\left(z\le \frac{\overline{x}-3.1}{0.05}\right)=0.85$

The z - value is obtained from the z - table as $z=1.04$

$1.04=\frac{\overline{x}-3.1}{0.05}$

$1.04\times 0.05=\overline{x}-3.1$

$0.052=\overline{x}-3.1$

$\overline{x}=3.1+0.052$

$\overline{x}=3.152$

Thus, the number of minutes required is 3.152.