1. MNM Corpor ation gives each of its employees an aptitude test. The scores on
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1. MNM Corpor ation gives each of its employees an aptitude test. The scores on the test have a mean of 60 and a standard deviation of 15. A points] simple random sample of 35 is taken from the population. [20 What is the expected value of the sampling distribution of x? loo b. What is the standard deviation of the sampling distribution of x? 1 5 cWhat is the shape of the sampling distribution of x? Why is it this shape? d. What is the random variable for the sampling distribution in this problem? Define it in words What is the probability that the average aptitude test score in the sample will be between 58.65 and 62.25? e. f. What is the probability that the average aptitude test score in the sample will be greater than 65.68? g. What is the probability that the average aptitude test score in the sample will be less than 58.69? h. Find the average aptitude test score that puts 2.5% of the averages above it.Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
mean of the sampling distribution ( x ) = 60
b.
standard Deviation ( sd )= 1.5/ Sqrt ( 35 ) =0.2535
sample size (n) = 35
c.
shape of the sampling distribution of x is bell shaped becuse given data clearly saying that it is normally distrubuted.
d.
X~N(60, 225)
e.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 58.65) = (58.65-60)/1.5/ Sqrt ( 35 )
= -1.35/0.25354628
= -5.3244718
= P ( Z <-5.3244718) From Standard Normal Table
= 0.00000005
P(X < 62.25) = (62.25-60)/1.5/ Sqrt ( 35 )
= 2.25/0.25354628 = 8.87411967
= P ( Z <8.87411967) From Standard Normal Table
= 1
P(58.65 < X < 62.25) = 1-0.00000005 = 0.99999995
f.
P(X > 65.68) = (65.68-60)/1.5/ Sqrt ( 35 )
= 5.68/0.254= 22.4022
= P ( Z >22.4022) From Standard Normal Table
= 0
g.
P(X < 58.69) = (58.69-60)/1.5/ Sqrt ( 35 )
= -1.31/0.2535= -5.1667
= P ( Z <-5.1667) From Standard NOrmal Table
= 0.00000012
h.
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.959964
P( x-u / (s.d) > x - 60/15) = 0.025
That is, ( x - 60/15) = 1.959964
--> x = 1.959964 * 15+60 = 89.39946
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