Question

Given a population with a mean of *µ* = 100 and a
variance *σ*^{2} = 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

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