Jamaica Auto Service is a popular place for car service. It is also known that the mean time taken for Oervice at this station is 15 minutes per car and the standard deviation is 2.5 minutes. It is also known that the time taken for service on a car follows a normal distribution. a) What is the probability that a randomly selected customer will have to wait at most 14 minutes for service? (round to four decimal places) b) What percentage of customers will have to wait between 12 and 18 minutes for their service? (round to nearest percentage) c) Is it likely that it may take longer than 23 minutes for service? Find the likelihood of happening this and justify your answer.

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**Jamaica Auto Service Statistical Analysis**

Jamaica Auto Service is a popular place for car service. It is also known that the mean time taken for service at this station is 15 minutes per car and the standard deviation is 2.5 minutes. The time taken for service on a car follows a normal distribution.

**Questions:**

a) What is the probability that a randomly selected customer will have to wait at most 14 minutes for service? *(round to four decimal places)*

b) What percentage of customers will have to wait between 12 and 18 minutes for their service? *(round to nearest percentage)*

c) Is it likely that it may take longer than 23 minutes for service? Find the likelihood of this happening and justify your answer.

**Instructions:**

To solve these questions, you will need to use your knowledge of the properties of the normal distribution and perhaps a standard normal distribution table (Z-table).

Remember:

- To find probabilities related to the normal distribution, convert the given times to Z-scores using the formula:

  \[ Z = \frac{(X - \mu)}{\sigma} \]

  where \( \mu \) is the mean and \( \sigma \) is the standard deviation.

- Use the Z-scores to find corresponding probabilities from a Z-table.

**Example Calculation:**

If you want to find the probability that a randomly selected customer will have to wait less than or equal to 14 minutes, compute the Z-score for X = 14 minutes:

\[ Z = \frac{(14 - 15)}{2.5} = -0.4 \]

Look up -0.4 in the Z-table to find the cumulative probability.

We encourage you to try these calculations to enhance your statistical skills!
Transcribed Image Text:**Jamaica Auto Service Statistical Analysis** Jamaica Auto Service is a popular place for car service. It is also known that the mean time taken for service at this station is 15 minutes per car and the standard deviation is 2.5 minutes. The time taken for service on a car follows a normal distribution. **Questions:** a) What is the probability that a randomly selected customer will have to wait at most 14 minutes for service? *(round to four decimal places)* b) What percentage of customers will have to wait between 12 and 18 minutes for their service? *(round to nearest percentage)* c) Is it likely that it may take longer than 23 minutes for service? Find the likelihood of this happening and justify your answer. **Instructions:** To solve these questions, you will need to use your knowledge of the properties of the normal distribution and perhaps a standard normal distribution table (Z-table). Remember: - To find probabilities related to the normal distribution, convert the given times to Z-scores using the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where \( \mu \) is the mean and \( \sigma \) is the standard deviation. - Use the Z-scores to find corresponding probabilities from a Z-table. **Example Calculation:** If you want to find the probability that a randomly selected customer will have to wait less than or equal to 14 minutes, compute the Z-score for X = 14 minutes: \[ Z = \frac{(14 - 15)}{2.5} = -0.4 \] Look up -0.4 in the Z-table to find the cumulative probability. We encourage you to try these calculations to enhance your statistical skills!
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