Question

Personnel tests are designed to test a job​ applicant's cognitive​ and/or physical abilities. A particular dexterity test is administered nationwide by a private testing service. It is known that for all tests administered last​ year, the distribution of scores was approximately normal with mean 79 and standard deviation 6.6. a. A particular employer requires job candidates to score at least 83 on the dexterity test. Approximately what percentage of the test scores during the past year exceeded 83​? b. The testing service reported to a particular employer that one of its job​ candidate's scores fell at the 90th percentile of the distribution​ (i.e., approximately 90​% of the scores were lower than the​ candidate's, and only 10​% were​ higher). What was the​ candidate's score? _______________________________________________________________________ a. Approximately 27.1of the test scores during the past year exceeded 83.​(Round to one decimal place as​ needed.) A answer was 27.1 b.What was the​ candidate's score?__________________(Round to the nearest whole number as​ needed.) B answer is _______________

Answer 1

for normal distribution z score =(X-μ)/σ
here mean=       μ=	79
std deviation   =σ=	6.6000

a)

percentage of the test scores during the past year exceeded 83 :

probability =

P(X>83)

=

P(Z>0.61)=

1-P(Z<0.61)=

1-0.729=

0.271~27.1%

b)

for 90th percentile z score =1.28

hence the​ candidate's score =mean+z*std deviation=79+1.28*6.6=87.0 ( please try 88 if this comes wrong)