Question

Given a population with a mean of µ = 100 and a variance σ² = 12, assume the central limit theorem applies when the sample size is n ≥ 25. A random sample of size n = 52 is obtained. What is the probability that   98.00 < x < 100.76?

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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 of the sampling distribution ( x ) = 100
standard Deviation ( sd )= 12/ Sqrt ( 52 ) =1.6641
sample size (n) = 52

the probability that 98.00 < x < 100.76
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 98) = (98-100)/12/ Sqrt ( 52 )
= -2/1.6641
= -1.2019
= P ( Z <-1.2019) From Standard Normal Table
= 0.1147
P(X < 100.76) = (100.76-100)/12/ Sqrt ( 52 )
= 0.76/1.6641 = 0.4567
= P ( Z <0.4567) From Standard Normal Table
= 0.6761
P(98 < X < 100.76) = 0.6761-0.1147 = 0.5613