Respuesta :
Answer and explanation:
Given : Foot locker uses sales per square foot as a measure of store productivity. sales are currently running at an annual rate of $406 per square foot. you have been asked by management to conduct a study of a sample of 64 foot locker stores.Assume the standard deviate in annual sales per square foot for the population of all 3400 Foot locker stores is $80.
To find :
a) Show the sampling distribution of x bar, the sample mean annual sales per square foot for a sample of 64 Foot locker stores.
By the central limit theorem also specifies that [tex]\bar{x}[/tex] will have the same expected value as the population mean, that is [tex]E(\bar{x}) = 406[/tex]
The standard deviation of [tex]\bar{x}[/tex] is given by,
[tex]\sigma_{\bar{x}}=\frac{\sigma}{\sqrt n}[/tex]
[tex]\sigma_{\bar{x}}=\frac{80}{\sqrt {64}}[/tex]
[tex]\sigma_{\bar{x}}=\frac{80}{8}[/tex]
[tex]\sigma_{\bar{x}}=10[/tex]
b) What is the probability that the sample mean will be within $15 of the population mean?
The probability is given by, [tex]P(-15 <x-\bar{x} <15)[/tex]
[tex]P(\frac{-15}{\sigma_{\bar{x}}} <\frac{x-\bar{x}}{\sigma_{\bar{x}}} <\frac{15}{\sigma_{\bar{x}}})[/tex]
[tex]P(\frac{-15}{10} <\frac{x-\bar{x}}{\sigma_{\bar{x}}} <\frac{15}{10})[/tex]
[tex]P(-1.5<Z<1.5)[/tex]
From standard normal table,
[tex]P(-1.5<Z<1.5)=0.8664[/tex]
c) Suppose you find a sample mean of $380. what is the probability of finding a sample mean of $380 or less? would you consider such a sample to be unusually low performing group of stores?
[tex]P(\bar{x}<380)=P(Z<\frac{x-\bar{x}}{\sigma_{\bar{x}}})[/tex]
[tex]P(\bar{x}<380)=P(Z<\frac{380-406}{10})[/tex]
[tex]P(\bar{x}<380)=P(Z<\frac{-26}{10})[/tex]
[tex]P(\bar{x}<380)=P(Z<-2.6)[/tex]
From standard normal table,
[tex]P(\bar{x}<380)=0.0047[/tex]
Probabilities are used to determine the chance of an event.
- The sampling distribution of x are: 406 and 10
- The probability that the sample mean will be within $15 of the population mean is 0.86639
- The probability of finding a sample mean of $380 or less is 0.0046612
- A sample mean of $380 should not be considered
The given parameters are:
[tex]\mathbf{n = 64}[/tex]
[tex]\mathbf{\sigma = 80}[/tex]
[tex]\mathbf{mu = 406}[/tex]
(a) The sampling distribution of [tex]\mathbf{\bar x}[/tex]
The sample mean is an estimate of the population.
So,
[tex]\mathbf{E(x) =mu = 406}[/tex]
The standard deviation is:
[tex]\mathbf{\sigma_x = \frac{\sigma}{\sqrt n}}[/tex]
[tex]\mathbf{\sigma_x = \frac{80}{\sqrt {64}}}[/tex]
[tex]\mathbf{\sigma_x = \frac{80}{8}}[/tex]
[tex]\mathbf{\sigma_x = 10}[/tex]
The sampling distribution of x are: 406 and 10
(b) The probability that the sample mean will be within $15 of the population mean.
This is represented as:
[tex]\mathbf{\mu-15 \le P(x) \le \mu+15}\\[/tex]
Start by calculating the z-score using:
[tex]\mathbf{z = \frac{x -\mu}{\sigma_x}}[/tex]
When x = 15
[tex]\mathbf{z = \frac{15}{10} = 1.5}[/tex]
When x = -15
[tex]\mathbf{z = \frac{-15}{10} = -1.5}[/tex]
So, the probability becomes
[tex]\mathbf{\mu-15 \le P(x) \le \mu+15 = -1.5 \le Pz \le P(15)}[/tex]
Using z-score of probabilities, we have:
[tex]\mathbf{\mu-15 \le P(x) \le \mu+15 = 0.86639}[/tex]
The probability that the sample mean will be within $15 of the population mean is 0.86639
(c) Probability of finding a sample mean of $380 or less
This means that x = 380
Calculate the z-score
[tex]\mathbf{z = \frac{x -\mu}{\sigma_x}}[/tex]
[tex]\mathbf{z = \frac{380 -406}{10}}[/tex]
[tex]\mathbf{z = \frac{-26}{10}}[/tex]
[tex]\mathbf{z = -2.6}[/tex]
The required probability is:
[tex]\mathbf{P(x \le 380) = P(z \le -2.6)}[/tex]
Using z-score of probabilities, we have:
[tex]\mathbf{P(x \le 380) = 0.0046612}[/tex]
The probability of finding a sample mean of $380 or less is 0.0046612
A sample mean of $380 should not be considered, because the probability is much lesser than the normal sample mean
Read more about normal probabilities at:
https://brainly.com/question/16087516
