Suppose a normal distribution has a mean of 62 and a standard deviation of 4. What is the probability that a data value is between 57 and 65? Round your answer to the nearest tenth of a percent.A.69.8%B.67.8%C.68.8%D.66.8%SUBMITarrow_backPREVIOUS

To solve this problem, we need to convert the raw scores to z-scores and then look up the corresponding probabilities in the z-table.

Step 1: Convert the raw scores to z-scores. The formula for converting a raw score to a z-score is: z = (X - μ) / σ, where X is the raw score, μ is the mean, and σ is the standard deviation.

For X = 57, z = (57 - 62) / 4 = -1.25 For X = 65, z = (65 - 62) / 4 = 0.75

Step 2: Look up the corresponding probabilities in the z-table. The probability corresponding to z = -1.25 is 0.1056, which means there is a 10.56% chance that a data value is less than 57. The probability corresponding to z = 0.75 is 0.7734, which means there is a 77.34% chance that a data value is less than 65.

Step 3: Subtract the two probabilities to find the probability that a data value is between 57 and 65. P(57 < X < 65) = P(X < 65) - P(X < 57) = 0.7734 - 0.1056 = 0.6678 or 66.78%

So, the answer is D. 66.8%.