Question

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 8.1 ounces and standard deviation 0.19 ounces. For a SRS of 5 of these chocolate bars, what is the level L such that there is a 1% chance that the average weight is less than L?

Answer 1

To find the level L such that there is a 1% chance that the average weight is less than L, we need to determine the value of L from the standard normal distribution. We can find the critical value t* from the t-distribution table corresponding to a 1% one-tailed confidence level and n-1 degrees of freedom. Finally, we can use a t-distribution calculator or table to find the critical value t* and compute the value of L.

Step-by-step explanation:

To find the level L such that there is a 1% chance that the average weight is less than L, we need to determine the value of L from the standard normal distribution.

Since the sample size is small (n=5), we need to use a t-distribution with n-1 degrees of freedom. We can find the critical value t* from the t-distribution table corresponding to a 1% one-tailed confidence level and n-1 degrees of freedom.

Once we have the critical value t*, we can calculate L by multiplying it with the standard deviation of the sample (0.19) and dividing it by the square root of the sample size (sqrt(5)). The formula is L = mean - t* * (standard deviation / sqrt(sample size)).

Substituting the values, we get L = 8.1 - t* * (0.19 / sqrt(5)).

Finally, we can use a t-distribution calculator or table to find the critical value t* and compute the value of L.