Answer: 0.9746
Step-by-step explanation:
Given : A normally distributed population has a mean of 250 pounds and a standard deviation of 10 pounds.
i.e. [tex]\mu=250[/tex] [tex]\sigma=10[/tex]
Sample size : n= 20
Let [tex]\overline{x}[/tex] sample mean values.
Then, the probability that this sample will have a mean value between 245 and 255 be
[tex]P(245<\overline{x}<255) =P(\dfrac{245-250}{\dfrac{10}{\sqrt{20}}}<\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{255-250}{\dfrac{10}{\sqrt{20}}})\\\\=P(-2.236<z<2.236)\ \ [\because \ z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<2.236)-P(z<-2.236)\\\\=P(z<2.236)-(1-P(z<2.236))\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=2P(z<2.236)-1\\\\= 2( 0.9873)-1\ \ [\text{By z-table}]\\\\=0.9746[/tex]
Hence , the probability that this sample will have a mean value between 245 and 255 is 0.9746 .
