Question

A population has a mean of 200 and a standard deviation of 50. Suppose a random sample of 100 people is selected from this population. What is the probability that the sample mean will be within /- 5 of the population mean

Answer 1

Complete Question

A population has a mean of 200 and a standard deviation of 50. Suppose a random sample of 100 people is selected from this population. What is the probability that the sample mean will be within
`\pm 5` of the population mean

Answer:

The value is
`P(195 <  X  <  205 ) =  0.6827`

Explanation:

From the question we are told that

The mean is
`\mu = 200`

The population standard deviation is
`\sigma = 50`

The sample size is
`n = 100`

Generally the standard error of sample mean is mathematically represented as

`s = ( \sigma )/(√(n) )`

=>
`s = (50 )/(√(100) )`

=>
`s =5`

Generally the limits of
`\pm 5` within the population mean is mathematically represented as

`a = \mu - 5`

=>
`a = 200 - 5`

=>
`a = 195`

and

`b = \mu + 5`

=>
`b = 200 + 5`

=>
`b = 205`

Generally the probability that the sample mean will be within
`\pm 5` of the population mean is mathematically represented as

`P(a < X < b ) = P((a - \mu )/(s) < (X - \mu )/(s) < (b - \mu )/(s) )`

=>
`P(195 <  X  <  205 ) =  P(( 195-  200 )/(5) < (X -  \mu )/(s) < (205 -  200 )/(5)  )`

`(X -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \ X )`

=>
`P(195 <  X  <  205 ) =  P(-1< Z<1 )`

=>
`P(195 <  X  <  205 ) =  P(Z <  1)  -  P(Z<-1 )`

Generally from the z -table the probability of (Z < 1) and (Z<-1 ) is

`P(Z < 1)= 0.84134`

and

`P(Z<-1 )   = 0.15866`

So

`P(195 <  X  <  205 ) =  0.84134  -  0.15866`

`P(195 <  X  <  205 ) =  0.6827`