A population has a mean of 300 and a standard deviation of 70, Suppose a sample

Question

A population has a mean of 300 and a standard deviation of 70, Suppose a sample of size 100 is selected and is used to estimate . use-table. a. what is the probability that the sample mean will be within +/ 4 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.) *q-please shar merov to use ln excel b. What is the probability that the sample mean will be within- 18 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places) 0.99

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
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 ( u ) = 300
standard Deviation ( sd )= 70
sample size (n) = 100
a.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 296) = (296-300)/70/ Sqrt ( 100 )
= -4/7
= -0.5714
= P ( Z <-0.5714) From Standard Normal Table
= 0.2839
P(X < 304) = (304-300)/70/ Sqrt ( 100 )
= 4/7 = 0.5714
= P ( Z <0.5714) From Standard Normal Table
= 0.7161
P(296 < X < 304) = 0.7161-0.2839 = 0.4323

We use excel formula to reach the solution
= ROUND(NORMDIST(296,300,70/SQRT(70),TRUE)-NORMDIST(304,300,70/SQRT(70),TRUE),4))

b.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 282) = (282-300)/70/ Sqrt ( 100 )
= -18/7
= -2.5714
= P ( Z <-2.5714) From Standard Normal Table
= 0.0051
P(X < 318) = (318-300)/70/ Sqrt ( 100 )
= 18/7 = 2.5714
= P ( Z <2.5714) From Standard Normal Table
= 0.9949
P(282 < X < 318) = 0.9949-0.0051 = 0.9899

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