Suppose that the test scores of students are normally distributed with the mean 76 and the variance of 64. The probability of students scoring between 70 and 82 is?

Question

0.7500


0.2266


0.7734


0.5468


0.6124

Answer 1

To find the probability of students scoring between 70 and 82, we need to calculate the z-scores for each score and then use the standard normal distribution table.

First, calculate the z-score for 70:
z = (x - mean) / standard deviation
z = (70 - 76) / √64
z = -6 / 8
z = -0.75

Next, calculate the z-score for 82:
z = (82 - 76) / √64
z = 6 / 8
z = 0.75

Now, look up the z-scores in the standard normal distribution table to find the corresponding probabilities:
P(z < -0.75) = 0.2266
P(z < 0.75) = 0.7734

To find the probability of students scoring between 70 and 82:
P(70 < x < 82) = P(x < 82) - P(x < 70)
P(70 < x < 82) = 0.7734 - 0.2266
P(70 < x < 82) = 0.5468

Therefore, the probability of students scoring between 70 and 82 is 0.5468.

So, the correct option is:
0.5468