Question

Scores on previous driver's tests taken by 16-year-olds were normally distributed with a mean of 82 and a standard deviation of 3.1. George will be taking the driving test tomorrow, what is the probability that he will receive at least an 88.2 on the test?

a. 0.0038
b. 0.0192
c. 0.0384
d. 0.1920

Answer 1

The probability that George will score at least an 88.2 on his driver's test is approximately 0.0228, which is not exactly reflected in any of the provided answer choices. The closest answer choice is 0.0384, but it should be noted that this is only due to answer constraints.

Step-by-step explanation:

You want to determine the probability that George will score at least an 88.2 on his driver's test. To do this, we will use the Z-score formula which is Z = (X - μ) / σ, where X is the score of interest, μ is the mean, and σ is the standard deviation.

First, calculate the Z-score for George's desired score of 88.2:

Z = (88.2 - 82) / 3.1 = 2

Next, look up the Z-score in the standard normal distribution table or use a calculator to find the probability associated with a Z-score of 2, which gives us the area to the left under the curve. Remember that George's score needs to be at least 88.2, so we're looking for the area to the right of the Z-score of 2.

The total area under the normal curve is 1, so the area to the right is 1 minus the area to the left. The area to the left for a Z-score of 2 is approximately 0.9772. Therefore, the probability that George will score at least 88.2 is:

1 - 0.9772 = 0.0228.

The closest available answer to 0.0228 is option c, 0.0384, but this is just an approximation due to limitations in the available answer choices.

Answer 2

The probability of George receiving at least an 88.2 on his driving test is calculated by determining the z-score and referencing the standard normal distribution. His z-score is 2, which corresponds to a probability of Z < 2 being 0.9772. Subtracting from 1, we get a probability of 0.0228, which does not match the options provided, indicating a possible error in the question or options.

Step-by-step explanation:

To find the probability that George will receive at least an 88.2 on his driving test, we need to calculate the z-score and then determine the probability associated with this z-score using the standard normal distribution table. First, we calculate the z-score using the formula:

z = (X - μ) / σ

Where X is George's score, μ is the mean score, and σ is the standard deviation. Plugging in the values, we get:

z = (88.2 - 82) / 3.1 = 2

Using a standard normal distribution table or a calculator, we can find the probability of getting a z-score less than 2 is about 0.9772. Since we want the probability of scoring at least 88.2, we subtract this from 1:

P(X ≥ 88.2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228

This is not one of the options provided, which suggests there might be a typo or miscalculation. However, from the options available, the closest answer would be option (d) 0.1920 which is incorrect based on our calculation.