Question

Suppose that X is normally distributed with mean 85 and standard deviation 11. a. What proportion of X is greater than 102.6? (Provide answer rounded to four decimal places.)

Answer 1

The proportion of X that is greater than 102.6 in a normal distribution with a mean of 85 and a standard deviation of 11 is approximately 0.0548, or 5.48%.

Step-by-step explanation:

To calculate the proportion of X that is greater than 102.6 when X is normally distributed with a mean (μ) of 85 and a standard deviation (σ) of 11, we first need to find the z-score for 102.6. The z-score can be found using the formula:

Z = (X - μ) / σ

Where X is the value we're comparing to the mean, μ is the mean of the distribution, and σ is the standard deviation. Plugging our values in, we get:

Z = (102.6 - 85) / 11 = 1.6

Next, we use a standard normal distribution table or a calculator to find the probability that Z is greater than 1.6. The table or calculator tells us that the proportion less than 1.6 is approximately 0.9452. To find the proportion greater than 102.6, we subtract this from 1:

1 - 0.9452 = 0.0548

Therefore, the proportion of X greater than 102.6 is approximately 0.0548, or 5.48%

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