Question

Agency leadership is interested in analyzing the engine sizes of this sample of 750 vehicles.​ (Use the mean and standard deviation of the Engine Size​ (L) data.​ Also, if appropriate based upon your visual analysis of a histogram of the Engine Size​ (L) data, use the Normal distribution to answer this​ question.) Calculate the probability of randomly selecting a vehicle with an engine size less than 2.7 L. enter your response here​% ​(Round to two decimal places as​ needed.) Part 2 Calculate the probability of randomly selecting a vehicle with an engine size greater than 3.9 L. enter your response here​% ​(Round to two decimal places as​ needed.) Part 3 Calculate the probability of randomly selecting a vehicle with an engine between than 3.1 L and 4.2 L. enter your response here​% ​(Round to two decimal places as​ needed.) Part 4 Calculate the engine size that represents the 10th percentile of this sample. enter your response here L ​(Round to two decimal places as​ needed.)

Answer 1

The engine size follows a normal distribution, and each probability is dependent on the z-scores

The probability of that a vehicle has an engine size less than 2.7 L.

The given parameters are:

• Mean = 3.60
• Standard deviation = 0.48

Calculate the z-score using:

$z = \frac{x - \mu}{\sigma}$

This gives

$z = \frac{2.7 - 3.60}{0.48}$

Evaluate

z = -1.875

The probability is then calculated using

P(x < 2.7) = P(z < -1.875)

Using the z table of probabilities, we have:

P(x < 2.7) = 0.030396

Express as percentage

P(x < 2.7) = 3.0396%

Approximate

P(x < 2.7) = 3.04%

Hence, the probability of randomly selecting a vehicle with an engine size less than 2.7 L is 3.04%

The probability that a vehicle has an engine size greater than 3.9 L

Calculate the z-score using:

$z = \frac{x - \mu}{\sigma}$

This gives

$z = \frac{3.9 - 3.60}{0.48}$

Evaluate

z = 0.625

The probability is then calculated using

P(x > 3.9) = P(z < 0.625)

Using the z table of probabilities, we have:

P(x > 3.9) = 0.73401

Express as percentage

P(x > 3.9) = 73.401%

Approximate

P(x > 3.9) = 73.40%

Hence, the probability of randomly selecting a vehicle with an engine size greater than 3.9 L is 73.40%

The probability that a vehicle has an engine between than 3.1 L and 4.2

Calculate the z-scores at x = 3.1 and 4.2 using:

$z = \frac{x - \mu}{\sigma}$

This gives

$z = \frac{3.1 - 3.60}{0.48} = -1.04$

$z = \frac{4.2 - 3.60}{0.48} = 1.25$

The probability is then calculated using

P(3.1 < x < 4.2) = P(-1.04 < z < 1.25)

This gives

P(3.1 < x < 4.2) = P(z < 1.25) - P(z < -1.04)

Using the z table of probabilities, we have:

P(3.1 < x < 4.2) = 0.89435 - 0.14917

Evaluate the difference

P(3.1 < x < 4.2) = 0.74518

Express as percentage

P(3.1 < x < 4.2) = 74.518%

Approximate

P(3.1 < x < 4.2) = 74.52%

Hence, the probability of randomly selecting a vehicle with an engine size between 3.1 L and 4.2 L is 74.52%

The engine size that represents the 10th percentile of this sample

At the 10th percentile, the z-score is:

z = -1.28

So, we have:

P = P(z > -1.28)

Using the z table of probabilities, we have:

P = 0.90

Hence, the engine size that represents the 10th percentile of this sample is 0.90 L

Read more about z-scores at:

brainly.com/question/25638875

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