Question

A school district requires all graduating seniors to take a mathematics test. This year, the test scores were approximately normally distributed for the 1208 graduating seniors. The mean score on the test was a 74 and the standard deviation was 11.

What percent of the graduating seniors had a test score above 85?

Answer 1

Answer:

hey mate

Step-by-step explanation:

teri mehfil m kismat azma kr hm bhi dekhenge. z = (X - Mean)/SD

z = (X - Mean)/SDz = (85 - 74)/11 = + 1

z = (X - Mean)/SDz = (85 - 74)/11 = + 1Required percent = P(X > 85)*100

z = (X - Mean)/SDz = (85 - 74)/11 = + 1Required percent = P(X > 85)*100= P(z > 1)*100

z = (X - Mean)/SDz = (85 - 74)/11 = + 1Required percent = P(X > 85)*100= P(z > 1)*100= [1 - P(z < 1)]*100

z = (X - Mean)/SDz = (85 - 74)/11 = + 1Required percent = P(X > 85)*100= P(z > 1)*100= [1 - P(z < 1)]*100= [1 - 0.8413]*100

z = (X - Mean)/SDz = (85 - 74)/11 = + 1Required percent = P(X > 85)*100= P(z > 1)*100= [1 - P(z < 1)]*100= [1 - 0.8413]*100= 0.1587*100

z = (X - Mean)/SDz = (85 - 74)/11 = + 1Required percent = P(X > 85)*100= P(z > 1)*100= [1 - P(z < 1)]*100= [1 - 0.8413]*100= 0.1587*100= 15.87% or 16% on rounding