In: Statistics and Probability
A sample of size 47 will be drawn from a population with mean 19 and standard deviation 14. Find the probability that will be greater than 22.
Solution :
Given that,
mean =
= 19
standard deviation =
=14
=
=19
=
/
n = 14/
47 = 2.04
P( > 22) = 1 - P(
< 22)
= 1 - P[(
-
) /
< (22-19) /2.04 ]
= 1 - P(z <1.47 )
Using z table
= 1 - 0.9292
= 0.0708
probability= 0.0708