Question

The distribution of the number of people in line at a grocery store has a mean of 3 and a variance of 9. A sample of the numbers of people in line in 50 stores is taken.

(a) Calculate the probability that the sample mean is more than 4? Round values to four decimal places.
(b) Calculate the probability the sample mean is less than 2.5. Round answers to four decimal places.
(c) Calculate the probability that the the sample mean differs from the population mean by less than 0.5. Round answers to four decimal places.

Answer 1

Answer:

a) $P(\bar X >4)=P(Z>\frac{4-3}{\frac{3}{\sqrt{50}}}=2.357)$

$P(Z>2.357)=1-P(Z<2.357) =1-0.9908=0.0092$

b) $P(\bar X <2.5)=P(Z>\frac{2.5-3}{\frac{3}{\sqrt{50}}}=-1.179)$

$P(Z<-1.179)=0.1192$

c)  

$P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}} <Z<\frac{3.5-3}{\frac{3}{\sqrt{50}}})$

$P(-1.179<Z<1.179)=P(Z<1.179)-P(Z<-1.179)=0.8808-0.1192=0.7616$Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the number of people of a population, and for this case we know that:

Where $\mu=3$ and $\sigma=\sqrt{9}=3$

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".

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

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

And we want to find this probability:

$P(\bar X >4)= P(z> \frac{4-3}{\frac{3}{\sqrt{50}}})$

And using a calculator, excel or the normal standard table we have that:

$P(Z>2.357)=1-P(Z<2.357) =1-0.9908=0.0092$

Part b

$P(\bar X <2.5)= P(z> \frac{2.5-3}{\frac{3}{\sqrt{50}}})$

And using a calculator, excel or the normal standard table we have that:

$P(Z<-1.179)=0.1192$

Part c

For this case we want this probability:

$P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}} <Z<\frac{3.5-3}{\frac{3}{\sqrt{50}}})$

And using a calculator, excel or the normal standard table we have that:

$P(-1.179<Z<1.179)=P(Z<1.179)-P(Z<-1.179)=0.8808-0.1192=0.7616$