Question

5 multiple choice 10 points
the distribution of long - jump distance in a statewide high school competition held at ohs is approximately normal with a mean of 18.5 feet and a standard deviation of 3.2 feet. two students, brooke and satchel, from ohs competed in the long - jump. satchels distance was.52 standard deviation below the mean of the distribution. brookes distance was at the 48th percentile of the distribution.
at approximately what percentile was the distance jumped by satchel, and how did his distance compare to brookes distance?
satchels distance was at the 30th percentile of the distribution and was less than brookes distance.
satchels distance was at the 48th percentile of the distribution and was equal to brookes distance.
satchel aint really having it!! shes still tight that mr. r and the class trolled her thinking she forgot to take the exam.
satchels distance was at the 48th percentile of the distribution and was less than brookes distance.
satchels distance was at the 52th percentile of the distribution and was greater than brookes distance.

Explanation:

Step1: Recall the z - score and percentile relationship

The z - score represents the number of standard deviations a data - point is from the mean. If a value has a z - score of \(z\), we can use the standard normal distribution table (z - table) to find the corresponding percentile. Given that Satchel's distance was \(z=- 0.52\) standard deviation below the mean.

Step2: Use the z - table

Looking up the z - score of \(z =-0.52\) in the standard normal distribution table, the area to the left of \(z=-0.52\) (which represents the percentile) is approximately \(0.3015\), or about the 30th percentile.

Step3: Compare with Brooke's percentile

Brooke's distance was at the 48th percentile. Since \(30\lt48\), Satchel's distance was less than Brooke's distance.

Answer:

Satchel's distance was at the 30th percentile of the distribution and was less than Brooke's distance.