pts) Given a normal population whose mean is 635 and whose standard deviation is
ID: 2922535 • Letter: P
Question
pts)
Given a normal population whose mean is 635 and whose standard deviation is 21, find each of the following:
A. The probability that a random sample of 6 has a mean between 637 and 643.
Probability =
B. The probability that a random sample of 14 has a mean between 637 and 643.
Probability =
C. The probability that a random sample of 22 has a mean between 637 and 643.
Probability =
A sample of n=22 observations is drawn from a normal population with =950 and =210. Find each of the following:
A. P(X¯>1048)
Probability =
B. P(X¯<873)
Probability =
C. P(X¯>891)
Probability =
Explanation / Answer
Q1.
NORMAL DISTRIBUTION
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)
mean ( u ) = 635
standard Deviation ( sd )= 21
a.
sample size (n) = 6
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 637) = (637-635)/21/ Sqrt ( 6 )
= 2/8.5732
= 0.2333
= P ( Z <0.2333) From Standard Normal Table
= 0.5922
P(X < 643) = (643-635)/21/ Sqrt ( 6 )
= 8/8.5732 = 0.9331
= P ( Z <0.9331) From Standard Normal Table
= 0.8246
P(637 < X < 643) = 0.8246-0.5922 = 0.2324
b.
sample size (n) = 14
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 637) = (637-635)/21/ Sqrt ( 14 )
= 2/5.6125
= 0.3563
= P ( Z <0.3563) From Standard Normal Table
= 0.6392
P(X < 643) = (643-635)/21/ Sqrt ( 14 )
= 8/5.6125 = 1.4254
= P ( Z <1.4254) From Standard Normal Table
= 0.923
P(637 < X < 643) = 0.923-0.6392 = 0.2838
c.
sample size (n) = 22
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 637) = (637-635)/21/ Sqrt ( 22 )
= 2/4.4772
= 0.4467
= P ( Z <0.4467) From Standard Normal Table
= 0.6725
P(X < 643) = (643-635)/21/ Sqrt ( 22 )
= 8/4.4772 = 1.7868
= P ( Z <1.7868) From Standard Normal Table
= 0.963
P(637 < X < 643) = 0.963-0.6725 = 0.2906
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