Question

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4. [20 Points! The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with a mean of 17 minutes and a standard deviation of 2 minutes. (a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take longer, the customer will receive the service for hall.price. What percent of customers receive the service for half-price? (b) If the automotive center does not want to give the discount to more than 2% of its customers, how long should it make the guaranteed time limit? 5. [10 Points Acconding to a study conducted by a statistical organization, the proportion of Americans who are satisfied with the way things are going in their lives is 0.8. Suppose that a random sample of 110 Americans is obtained. a. Describe the sampling distribution of sample proportion h. Using the distribution from part(s), what is the probability that at least 2 Americans in the comple are satisfied with their lives? 6. [20 Points) In a poll, 202 of 1010 randomly selected adults aged 18 or older stated that they believe there is too little spending on national defense. Use this information to complete parts (a) through (e) below a. Obtain a point estimate for the proportion of adults aged 18 or older who feel there is too little spending on national defense. Construct a 90% confidence interval for the proportion of adults aged 18 or older who believe there is too little spending on national defense.

Answer 1

Answer 4 :

Mean () = 17 min

Standard Deviation () = 2 min

Answer (a) :

P (X > 20 ) = 1- P(X < 20)

We know that :

Therefore,

Using the Standard Normal Distribution Table, we get:

1 - 0.9332

= 0.0668

= 6.68%

Therefore, the Percentage of Customers that receive the Service rate for Half Price is 6.68%

Answer (b) :

We want to Restrict it to not more than 2%, i.e. P(X > x) = 0.02

Therefore, P(X < x) = 1- 0.02 = 0.98

and the Corresponding Z-Value is 2.054

Therefore,

Now, we just need to solve for X

4.108 = X - 19

X = 23.108 min.

Therefore, Guaranteed Time Limit Should be made 23.108 min.

Answer 5 :

Answer a and b :

Answer 6 :

Answer a :

proportion estimate for population proportion = sample proportion

=202/1010 = 0.20

i) Sample size is large

ii) 1010 were selected at random

Hence we assume proportion to follow normal distribution.

Yes. the requirements are satisfied

Answer b :

90% alpha = 1.645

Margin of error = 1.645* std error of proportion

= ±0.0207

STd error^2 = p(1-p)/n

Confidence interval at 90% = -0.1793 to 0.2207

= (-0.179, 0.221)

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