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 [tex]\pm 5[/tex] of the population mean
Answer:
The value is [tex]P(195 < X < 205 ) = 0.6827 [/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 200[/tex]
The population standard deviation is [tex]\sigma = 50[/tex]
The sample size is [tex]n = 100[/tex]
Generally the standard error of sample mean is mathematically represented as
[tex]s = \frac{ \sigma }{\sqrt{n} }[/tex]
=> [tex]s = \frac{50 }{\sqrt{100} }[/tex]
=> [tex]s =5[/tex]
Generally the limits of [tex]\pm 5[/tex] within the population mean is mathematically represented as
[tex]a = \mu - 5[/tex]
=> [tex]a = 200 - 5[/tex]
=> [tex]a = 195[/tex]
and
[tex]b = \mu + 5[/tex]
=>[tex]b = 200 + 5[/tex]
=>[tex]b = 205[/tex]
Generally the probability that the sample mean will be within [tex]\pm 5[/tex] of the population mean is mathematically represented as
[tex]P(a < X < b ) = P(\frac{a - \mu }{s} < \frac{X - \mu }{s} < \frac{b - \mu }{s} )[/tex]
=> [tex]P(195 < X < 205 ) = P(\frac{ 195- 200 }{5} < \frac{X - \mu }{s} < \frac{205 - 200 }{5} )[/tex]
[tex]\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )[/tex]
=> [tex]P(195 < X < 205 ) = P(-1< Z<1 )[/tex]
=> [tex]P(195 < X < 205 ) = P(Z < 1) - P(Z<-1 )[/tex]
Generally from the z -table the probability of (Z < 1) and (Z<-1 ) is
[tex]P(Z < 1)= 0.84134[/tex]
and
[tex] P(Z<-1 ) = 0.15866 [/tex]
So
[tex]P(195 < X < 205 ) = 0.84134 - 0.15866 [/tex]
[tex]P(195 < X < 205 ) = 0.6827 [/tex]
