The engine size follows a normal distribution, and each probability is dependent on the z-scores
<h3>The probability of that a vehicle has an engine size less than 2.7 L.</h3>
The given parameters are:
- Mean = 3.60
- Standard deviation = 0.48
Calculate the z-score using:
This gives
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%
<h3>The probability that a vehicle has an engine size greater than 3.9 L</h3>
Calculate the z-score using:
This gives
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%
<h3>The probability that a vehicle has an engine between than 3.1 L and 4.2 </h3>
Calculate the z-scores at x = 3.1 and 4.2 using:
This gives
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%
<h3>The engine size that represents the 10th percentile of this sample</h3>
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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