Question

The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $5.00. Assume the distribution of amounts purchased follows the normal distribution. For a sample of 50 customers, answer the following questions.
(a) What is the likelihood the sample mean is at least $25.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
(c) Within what limits will 90 percent of the sample means occur? (Round your answers to 2 decimal places.)

Answer 1

Answer:

a. $\mathtt{P(X \geq 25) =0.0170}$     ( to four decimal places)

b. $P(22.5<X<25) = 0.9043$   ( to four decimal places )

c. The limits will be between the interval of   ( 22.33,24.67 )

Step-by-step explanation:

Given that :

mean = 23.50

standard deviation = 5.00

sample size = 50

The objective is to calculate the following:

(a)  What is the likelihood the sample mean is at least $25.00?

Let X be the random variable, the probability that the sample mean is at least 25.00 is:

$P(X \geq 25) = 1 - P(\dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{25- 23.50}{ \dfrac{5}{\sqrt{ 50}} })$

$P(X \geq 25) = 1 - P(Z< \dfrac{1.5}{ \dfrac{5}{7.07107}} })$

$P(X \geq 25) = 1 - P(Z< \dfrac{1.5 \times 7.071}{ {5}})$

$P(X \geq 25) = 1 - P(Z< 2.1213)$

$P(X \geq 25) = 1 - P(Z< 2.12)$   to two decimal places

From the normal tables :

$P(X \geq 25) = 1 - 0.9830$

$\mathtt{P(X \geq 25) =0.0170}$     ( to four decimal places)

(b) What is the likelihood the sample mean is greater than $22.50 but less than $25.00?

$P(22.5<X<25) = P(\dfrac{X-\mu}{\dfrac{\sigma}{\sqrt{n}}} <\dfrac{25-23.5}{\dfrac{5}{\sqrt{50}}} ) - P(\dfrac{X-\mu}{\dfrac{\sigma}{\sqrt{n}}} <\dfrac{22.5-23.5}{\dfrac{5}{\sqrt{50}}} )$

$P(22.5<X<25) = P(Z<\dfrac{1.5}{\dfrac{5}{7.071}} ) - P(Z<\dfrac{-1}{\dfrac{5}{7.071}} )$

$P(22.5<X<25) = P(Z<2.12) - (Z<-1.41 )$

$P(22.5<X<25) = (0.9830 ) - (0.0787)$

$P(22.5<X<25) = 0.9043$  to four decimal places

(c) Within what limits will 90 percent of the sample means occur?

At 90 % confidence interval, level of significance = 1 - 0.90 = 0.10

The critical value for the $z_{\alpha/2} = 0.05$ = 1.65

Standard Error = $\dfrac{\sigma}{\sqrt{n}}$

Standard Error =  $\dfrac{5}{\sqrt{50}}$

Standard Error = 0.7071

Therefore, at 90 percent of the sample means, the limits will be between the intervals of : $(\mu \pm z_{\alpha/2} \times S.E)$

Lower limit =  ( 23.5 - (1.65×0.707) )

Lower limit =  ( 23.5 - 1.16655 )

Lower limit = 22.33345

Lower limit = 22.33    (to two decimal places).

Upper Limit = ( 23.5 + (1.65*0.707) )

Upper Limit = ( 23.5 + 1.16655 )

Upper Limit = 24.66655

Upper Limit = 24.67

The limits will be between the interval of   ( 22.33,24.67 )