Respuesta :
Answer:
Probability that the sample variance exceeds 3.10 is less than 0.05%.
Step-by-step explanation:
We are given that a process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of 1.75.
Also, a random sample of 20 of these batches is chosen.
The probability distribution that we will used here is of Chi-square distribution, i.e.;
[tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, [tex]s^{2}[/tex] = sample variance
[tex]\sigma^{2}[/tex] = population variance = 1.75
n = sample of batches = 20
So, probability that the sample variance exceeds 3.10 is given by = P([tex]s^{2}[/tex] > 3.10)
P([tex]s^{2}[/tex] > 3.10) = P( [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] > [tex]\frac{(20-1) \times 3.10^{2} }{1.75^{2} }[/tex] )
= P( [tex]\chi^{2} __2_0_-_1[/tex] > 59.62) = P( [tex]\chi^{2} __1_9[/tex] > 59.62) = Less than 0.05%
From chi-square table we can observe that Probability that [tex]\chi^{2} __1_9[/tex] is greater than is less than 0.05%.
So, the probability that the sample variance exceeds 3.10 is less than 0.05%.
