The heights of the 430 National Basketball Association...

The heights of the 430 National Basketball Association players were listed on team rosters at the start of the $2005-2006$ season. The heights of basketball players have an approximate normal distribution with mean, $\mu=79$ inches and a standard deviation, $\sigma=3.89$ inches. For each of the following heights, calculate the z-score and interpret it using complete sentences. a. 77 inches b. 85 inches c. If an NBA player reported his height had a z-score of $3.5,$ would you believe him? Explain your answer.

Transcript

00:01 In problem 68, given the mean and the standard deviation for the heights of basketball players, where the mean equals 79 inches and the standard deviation equals 3 .89 inches.

00:23 It's required to calculate the z score for the following heights.

00:30 For part a, we have the height which is denoted by x.

00:36 77 inches and for part b we want to calculate that z score for the height which equals 85 inches we can calculate the z score using this formula z equals x minus mu divided by sigma then for part a z equals 77 minus mu which is 79 divided by sigma then for part a z equals 77 sigma 3 .89 which equals negative 0 .514 and to interpret sentence using words we have the height which equals 77 inches is below the mean which equals 79 inches by 0 .514 standard deviations standard deviations then this height is below the mean by about 0 .5 the standard deviation half the standard division we repeat the same procedure for part b we calculate z as x x now is 85 minus mu divided by sigma which gives 1 .542 and to interpret this in a complete sentence this means the height which equals 85 inches is above or to the right of the mean above the mean which equals 79 inches by 1 .542 or 1 .5 standard divisions and for bar c one of the players reported that his height had a score of 3 .5.

03:19 His height had a z score of 3 .5.

03:26 Would we believe him? is that true? let's think of it as a probability.

03:38 What is the probability to have a z score of 3 .5? using the tables, we know that the probability to have z, the probability to have x, the probability to have x, the probability to have z, z score, which smaller than equals 3 .5 equals 99 .96...

Question

Please give Ace some feedback