A candy manufacturer must monitor deviations in the amount of sugar in their products They want their products to meet standards. They selected a random sample of 20 candies and found that the sandard deviation of that sample is 1.7. What is the probabilty of finding a sample variance as high or higher if the population variance is actually 3277 Assume the population distribution is normal.

To solve this problem, we can use the chi-squared distribution to test if the sample variance is significantly different from the population variance. The chi-squared distribution is commonly used to test the variances of two populations.

The test statistic for this problem is given by:

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

Where:
- \chi^{2} is the chi-squared value
- n is the sample size
- S^{2} is the sample variance
- \sigma^{2} is the population variance

In this case, we are interested in finding the probability of finding a sample variance as high or higher. This corresponds to the right-tail of the chi-squared distribution.

Step 1: Calculate the chi-squared value
Using the given values, we have:
n = 20
S^{2} = (1.7)^{2} = 2.89
\sigma^{2} = 3277

Plugging these values into the formula, we get:
\chi^2=\frac{(20-1)(2.89)}{3277}\approx0.0168

Step 2: Find the probability
To find the probability of finding a sample variance as high or higher, we need to find the right-tail probability of the chi-squared distribution.

The degrees of freedom for the chi-squared distribution in this case is n-1 = 19 .

Using a chi-squared table or a calculator, we can find that the right-tail probability for \chi^{2} = 0.0168 with df = 19 is approximately 0.9384.

Answer:
The probability of finding a sample variance as high or higher if the population variance is actually 3277 is approximately 1.