In: Statistics and Probability
A Population has a mean of 50 and a standard deviation of 15. If a random sample of 49 is taken, what is the probability that the sample mean is each of the following
a. greater than 54
b. less than 52
c. less than 47
d. between 45.5 and 51.5
e. between 50.3 and 51.3
Solution :
Given that ,
mean = = 50
standard deviation = = 15
n = 49
= 50 and
= / n = 15 / 49 = 2.1429
a.
P( > 54) = 1 - P( < 54)
= 1 - P(( - ) / < (54 -50) / 2.1429)
= 1 - P(z < 1.8666)
= 1 - 0.969
= 0.031
Probability = 0.031
b.
P( < 52) = P(( - ) / < (52 - 50) / 2.1429)
= P(z < 0.9333)
= 0.8247
Probability = 0.8247
c.
P( < 47) = P(( - ) / < (47 - 50) / 2.1429)
= P(z < -1.4000)
= 0.0808
Probability = 0.0808
d.
P( 45.5 < < 51.5) = P((45.5 - 50) / 2.1429)<( - ) / < (51.5 - 50) / 2.1429))
= P(-2.1000 < Z < 0.7000)
= P(Z < 0.7000) - P(Z < -2.1000)
= 0.758 - 0.0179
= 0.7401
Probability = 0.7401
e.
P(50.3 < < 51.3) = P((50.3 - 50) / 2.1429)<( - ) / < (51.3 - 50) / 2.1429))
= P(0.1400 < Z < 0.6067)
= P(Z < 0.6067) - P(Z < 0.1400)
= 0.728 - 0.5557
= 0.1723
Probability = 0.1723