Question

A normal population has a mean of 61 and a standard deviation of 4. You select a sample of 38.

Compute the probability that the sample mean is: (Round your z values to 2 decimal places and final answers to 4 decimal places.)

1. Less than 60.

2. Between 60 and 62.

3. Between 62 and 63.

4. Greater than 63.

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 ) = 61
standard deviation ( sd )= 4/ Sqrt ( 38 ) =0.6489
sample size (n) = 38
----------------------------------------------------------------------------------------
(a)
P(X < 60) = (60-61)/4/ Sqrt ( 38 )
= -1/0.6489= -1.5411
= P ( Z <-1.5411) From Standard NOrmal Table
= 0.0616
----------------------------------------------------------------------------------------
(b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 60) = (60-61)/4/ Sqrt ( 38 )
= -1/0.6489
= -1.5411
= P ( Z <-1.5411) From Standard Normal Table
= 0.0616
P(X < 62) = (62-61)/4/ Sqrt ( 38 )
= 1/0.6489 = 1.5411
= P ( Z <1.5411) From Standard Normal Table
= 0.9384
P(60 < X < 62) = 0.9384-0.0616 = 0.8767
----------------------------------------------------------------------------------------
(c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 62) = (62-61)/4/ Sqrt ( 38 )
= 1/0.6489
= 1.5411
= P ( Z <1.5411) From Standard Normal Table
= 0.9384
P(X < 63) = (63-61)/4/ Sqrt ( 38 )
= 2/0.6489 = 3.0822
= P ( Z <3.0822) From Standard Normal Table
= 0.999
P(62 < X < 63) = 0.999-0.9384 = 0.0606
----------------------------------------------------------------------------------------
(d)
P(X > 63) = (63-61)/4/ Sqrt ( 38 )
= 2/0.649= 3.0822
= P ( Z >3.0822) From Standard Normal Table
= 0.001