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san4es73 [151]
3 years ago
12

A process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of175. A ran

dom sample of 20 of these batches is chosen. Find the probability that the sample variance exceeds 3.10.
Mathematics
2 answers:
Tom [10]3 years ago
8 0

Answer:

the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

Step-by-step explanation:

since the variance  S² of the batch follows a normal distribution , then for a sample n of  20 distributions , then the random variable Z:

Z= S²*(n-1)/σ²

follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom

since

S² > 3.10 , σ²= 1.75 , n= 20

thus

Z > 33.65

then from χ² distribution tables:

P(Z > 33.65) = 0.02020

therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

Arisa [49]3 years ago
5 0

Answer:

P(s^2 >3.10) =P(\frac{(n-1)s^2}{\sigma^2}>\frac{19*3.10}{1.75})

P(chi^2_{19}>33.657)=1-P(\chi^2_{19]

Step-by-step explanation:

Previous concepts

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Data given and notation

For this case we can use the fact that the estimator of the population variance \sigma is the sample variance s^2, because E(s^2)=\sigma^2

The proof is this one:

Since E(\chi^2) = n-1 and

\chi^2 =\frac{(n-1) s^2}{\sigma^2}

When we take the expected value we got:

E[\frac{(n-1) s^2}{\sigma^2}]= n-1

E[s^2]=\frac{n-1}{n-1}\sigma^2

E[s^2]=\sigma^2

We have the distribution on this case given chi square.

Solution to the problem

The degrees of freddom on this case are given by

df=n-1=20-1=19

On this case we want this probability:

P(s^2 >3.10) =P(\frac{(n-1)s^2}{\sigma^2}>\frac{19*3.10}{1.75})

P(chi^2_{19}>33.657)=1-P(\chi^2_{19}

And we can use excel to find the probability with the following code:"=1-CHISQ.DIST(33.657,19,TRUE) "

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