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melisa1 [442]
3 years ago
10

The mean amount purchased by a typical customer at Churchill's Grocery Store is $23.50 with a standard deviation of $5.00. Assum

e 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.)
Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

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

b. P(22.5   ( 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

P(22.5

P(22.5

P(22.5

P(22.5  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 )

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