Answer :
Answer: 0.504772
Step-by-step explanation:
Given : The sugar content of the syrup is canned peaches is normally distributed. Assume the can is designed to have standard deviation [tex]\sigma=5[/tex] milligrams.
A random sample of n = 10 cans is studied.
Then, the sampling distribution of the sample variance is chi-square ([tex]\chi^2[/tex]) distribution witth [tex]n-1[/tex] degrees of freedom.
Sample standard deviation: [tex]s=4.8[/tex] milligrams.
Test statistic for chi-square =[tex]\chi^2=\dfrac{(n-1)s^2}{\sigma^2}[/tex]
[tex]\\\\=\dfrac{(10-1)(4.8)^2}{(5)^2}\\\\=8.2944[/tex]
By using the chi-square distribution table , the chance of observing the sample standard deviation greater than 4.8 milligrams will be :-
P-value = [tex]P(\chi^2>8.2944)=1-P(\chi^2\leq8.2944)[/tex]
[tex]=1-0.495228= 0.504772[/tex]
Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.504772