1.An appliance store sells major appliances (such as dishwashers, washing machines, dryers); for an additional fee, the store offers an extended service contract beyond the warranty period. The owner of this appliance store notes that she sold 145 dishwashers during the past 6 months; 45% of these customers also purchased the extended service contract. To maintain customer relations, the owner likes to contact her customers to see how they are enjoying their new appliances. She conducts a random sample of 35 of these customers. a)What is the probability that fewer than 40% of the customers in the sample purchased the service contract

Answer:

0.2776 = 27.76% probability that fewer than 40% of the customers in the sample purchased the service contract.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

45% of these customers also purchased the extended service contract.

This means that [tex]p = 0.45[/tex]

Sample of 35

This means that [tex]n = 35[/tex]

Mean and standard deviation:

[tex]\mu = 0.45[/tex]

[tex]s = \sqrt{\frac{0.45*0.55}{35}} = 0.0841[/tex]

What is the probability that fewer than 40% of the customers in the sample purchased the service contract?

This is the p-value of Z when X = 0.4. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.4 - 0.45}{0.0841}[/tex]

[tex]Z = -0.59[/tex]

[tex]Z = -0.59[/tex] has a p-value of 0.2776

0.2776 = 27.76% probability that fewer than 40% of the customers in the sample purchased the service contract.

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