Respuesta :
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.
Let X = distribution of test scores
SO, X ~ Normal([tex]\mu=72, \sigma^{2} =8.1^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean score = 72
[tex]\sigma[/tex] = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{78-72}{8.1}[/tex] ) = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{x-72}{8.1}[/tex] ) = 0.05
P(Z > [tex]\frac{x-72}{8.1}[/tex] ) = 0.05
Now, in the z table the critical value of x which represents the top 5% area is given as 1.645, i.e;
[tex]\frac{x-72}{8.1} = 1.645[/tex]
[tex]{x-72} = 1.645 \times 8.1[/tex]
x = 72 + 13.32 = 85.32
Hence, the candidate's score was 85.32.
