Solution :
Given that ,
mean =
= 75
standard deviation =
= 15
n = 36
=
= 75
=
/
n = 15 /
36 = 2.5
a) P(70.14 <
< 82.14)
= P[(70.14 - 75) / 2.5 < (
-
)
/
< (82.14 - 75) / 2.5)]
= P(-1.944 < Z < 2.856)
= P(Z < 2.856) - P(Z < -1.944)
Using z table,
= 0.9979 - 0.0259
= 0.9720
b) Using standard normal table,
P(Z
z) = 0.015
= 1 - P(Z
z) = 0.015
= P(Z
z ) = 1 - 0.015
= P(Z
2.17 ) = 0.985
z = 2.17
Using z-score formula
= z *
+
= 2.17 * 2.5 + 75
= 80.425
C = 80.43
P(
80.43) = 0.015
MNM Corporation gives each of its employees an aptitude test. The scores on the test...
5. MNM Corporation gives each of its employees an aptitude test. The scores on the test are normally distributed with a mean of 75 and a standard deviation of 15. A simple random sample of 25 is taken from a population of 500. a. What are the expected value, the standard deviation, and the shape of the sampling distribution of x? b. What is the probab]lity that the average aptitude test in the sample will be between 70.14 and 82.14?...
Business and Economic Statistics
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