Answer:
(a) 22.96% of the test scores during the past year exceeded 78.
(b) The candidate's score was 85.32.
Step-by-step explanation:
We are given that 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 72 and standard deviation 8.1.
<u><em>Let X = distribution of test scores</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean score = 72
= standard deviation = 8.1
(a) Now, percentage of the test scores during the past year which exceeded 78 is given by = P(X > 78)
P(X > 78) = P( >
) = P(Z > 0.74) = 1 - P(Z < 0.74)
= 1 - 0.7704 = 0.2296
The above probability is calculated by looking at the value of x = 0.74 in the z table which has an area of 0.77035.
Therefore, 22.96% of the test scores during the past year exceeded 78.
(b) Now, we given that the testing service reported to a particular employer that one of its job candidate's scores fell at the 95th percentile of the distribution and we have to find the candidate's score, that means;
P(X > x) = 0.05 {where x is the required candidate score}
P( >
) = 0.05
P(Z > ) = 0.05
<em>Now, in the z table the critical value of x which represents the top 5% area is given as 1.645, i.e;</em>
x = 72 + 13.32 = <u>85.32</u>
<u></u>
Hence, the candidate's score was 85.32.