Question
The mean score on a particular test is 79 and the standard deviation is 5.5. What is the probability that a randomly selected test score is more than 90?(1 point) Responses 34% 34% 68% 68% 5% 5% 95% 95%
Answers
GPT-4o mini
To find the probability that a randomly selected test score is more than 90 given that the mean score is 79 and the standard deviation is 5.5, we can use the properties of the normal distribution.
First, we need to calculate the z-score for a test score of 90:
\[
z = \frac{X - \mu}{\sigma}
\]
Where:
- \( X \) is the score (90 in this case),
- \( \mu \) is the mean (79),
- \( \sigma \) is the standard deviation (5.5).
Substituting the values:
\[
z = \frac{90 - 79}{5.5} = \frac{11}{5.5} = 2
\]
Now, we will use the z-score to find the probability. We can look up the z-score of 2 in a standard normal distribution table or use a calculator.
The z-score table or calculator will give us the probability of a z-score being less than 2. Typically, the probability for \( z < 2 \) is approximately 0.9772.
To find the probability that a score is more than 90, we subtract this value from 1:
\[
P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228
\]
This means that the probability that a randomly selected test score is more than 90 is approximately \( 0.0228 \) or 2.28%.
Since this value does not match any of the options provided, it appears that there may be an error in the choices given. However, based on standard probability calculations, the correct answer is approximately 2.28%.
If we round this to the nearest percentage among the options provided, it appears that none of the provided answers (34%, 68%, 5%, 95%) directly correspond, but the closest would be 5%.
First, we need to calculate the z-score for a test score of 90:
\[
z = \frac{X - \mu}{\sigma}
\]
Where:
- \( X \) is the score (90 in this case),
- \( \mu \) is the mean (79),
- \( \sigma \) is the standard deviation (5.5).
Substituting the values:
\[
z = \frac{90 - 79}{5.5} = \frac{11}{5.5} = 2
\]
Now, we will use the z-score to find the probability. We can look up the z-score of 2 in a standard normal distribution table or use a calculator.
The z-score table or calculator will give us the probability of a z-score being less than 2. Typically, the probability for \( z < 2 \) is approximately 0.9772.
To find the probability that a score is more than 90, we subtract this value from 1:
\[
P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228
\]
This means that the probability that a randomly selected test score is more than 90 is approximately \( 0.0228 \) or 2.28%.
Since this value does not match any of the options provided, it appears that there may be an error in the choices given. However, based on standard probability calculations, the correct answer is approximately 2.28%.
If we round this to the nearest percentage among the options provided, it appears that none of the provided answers (34%, 68%, 5%, 95%) directly correspond, but the closest would be 5%.