Question

A (very large) population has a mean µ of 800 and a standard deviation σ of 25. What is the probability that a sample mean x will be within ± 5 units of the population mean for each of the following sample sizes?

a. n = 50

b. n = 75

c. n = 100

Answer 1

Solution :

Given that,

mean = = 800

standard deviation = = 25

a)

n = 50

= = 800

= / n = 25 / 50 = 3.5355

P(795 < < 805) = 1 - (P((795 - 800) /3.5355 <( - ) / < (805 - 800) / 3.5355)))

=1 - ( P(-1.41 < Z <1.41))

= 1 - (P(Z < 1.41) - P(Z < -1.41)) Using standard normal table,  

= 1 - (0.9207 - 0.0793)

= 1 - 0.8414

Probability = 0.1586

b)

n = 75

= = 800

= / n = 25 / 75 = 2.8868

P(795 < < 805) = 1 - (P((795 - 800) /2.8868 <( - ) / < (805 - 800) / 2.8868)))

=1 - ( P(-1.73 < Z <1.73))

= 1 - (P(Z < 1.73) - P(Z < -1.73)) Using standard normal table,  

= 1 - (0.9582 - 0.0418)

= 1 - 0.9164

Probability = 0.0836

c)

n = 100

= = 800

= / n = 25 / 100 = 2.5

P(795 < < 805) = 1 - (P((795 - 800) /2.5 <( - ) / < (805 - 800) / 2.5)))

=1 - ( P(-2 < Z < 2))

= 1 - (P(Z < 2) - P(Z < -2)) Using standard normal table,  

= 1 - (0.9772 - 0.0228)

= 1 - 0.8414

Probability = 0.9544