Evan's home run hitting distance is normally distributed with a mean of 398 feet and a standard deviation of 28 feet. He wanted to find the probability that his home runs traveled at least 385 feet. He calculated the z-score to be −0.46 and looked up the probability on the Standard Normal Probabilities table. He found that the table stated his probability as 0.3228. Determine whether Evan made an error in his calculation and explain.

A. No, he did not make an error. The table contains incorrect information.
B. Yes, he made an error. He failed to add 1 to the probability, because he was seeking the probability that his home runs travel at least 385 feet. The correct probability is 1.6772.
C. No, he did not make an error. He looked up −0.46 on the table and the probability reads 0.3228.
D. Yes, he made an error. Because he was seeking the probability that his home runs traveled at least 385 feet, he looked up −0.46 and failed to subtract the probability from 1. The correct probability is 0.6772.
E. It cannot be determined whether Evan made an error.

Option D) Yes, he made an error. Because he was seeking the probability that his home runs traveled at least 385 feet, he looked up −0.46 and failed to subtract the probability from 1. The correct probability is 0.6772.

Step-by-step explanation:

We are given that Evan's home run hitting distance is normally distributed with a mean of 398 feet and a standard deviation of 28 feet i.e. = 398 feet and = 28 feet.

Also, He wanted to find the probability that his home runs  traveled at least 385 feet. He calculated the z-score to be −0.46 and looked up the probability on the Standard Normal Probabilities table. He found that the table stated his probability as 0.3228 .

The probability he stated is wrong because as we know that probability that his home runs traveled at least 385 feet is given by P(X >= 385 feet) ;

where, X = home run is being hit

            Z = ~ N(0,1)

P(X >= 385) = P( >= ) = P(Z >= -0.46) = P(Z <= 0.46) = 0.6772.

When looking at the z table we find that at 0.46 critical value of x, the probability area is 0.6772 .

a

step-by-step explanation: