# 1.A sample of 25 bottles is taken from the production line at a local bottling plant. Assume that the fill amounts follow a normal distribution.

The probability is 90% that the sample variance is less than what percent of the population variance?

2. The amount of money spent by tourists when visiting Disneyland is normally distributed with a standard deviation of \$43. How large of a sample must you take in order to ensure that the probability is 0.90 of getting a sample mean within \$10 of the population mean?

## 1. To find the probability that the sample variance is less than a certain percent of the population variance, we can use the Chi-square distribution.

First, we need to determine the degrees of freedom for the Chi-square distribution. In this case, since we have a sample size of 25, the degrees of freedom is 25 - 1 = 24.

Next, we need to find the chi-square value corresponding to a 90% probability (or a significance level of 0.10). The chi-square value can be obtained from a chi-square distribution table or calculated using statistical software. For a 90% probability and 24 degrees of freedom, the chi-square value is approximately 36.42.

Finally, we calculate the sample variance as ((n-1) * s^2) / chi-square, where n is the sample size (25), s^2 is the sample variance, and chi-square is the chi-square value.

Let's denote the sample variance as x. The inequality to solve is ((25-1) * x) / 36.42 < 1, where 1 represents 100% of the population variance.

Simplifying the inequality, we get 24x < 36.42.

Dividing both sides by 24, we get x < 1.5175.

Therefore, the probability is 90% that the sample variance is less than 1.5175% of the population variance.

2. To determine the sample size required to ensure a probability of 0.90 (or 90%) of getting a sample mean within \$10 of the population mean, we can use the Central Limit Theorem and the formula for the margin of error.

The Central Limit Theorem states that as the sample size increases, the sampling distribution of the sample mean approaches a normal distribution regardless of the shape of the population distribution.

The margin of error is calculated as z * (standard deviation / sqrt(n)), where z is the z-score representing the desired level of confidence, standard deviation is the population standard deviation (\$43 in this case), and n is the sample size.

We want the margin of error to be \$10, and we want a 90% probability, so the z-score corresponding to a 90% probability (or a significance level of 0.10) is approximately 1.645.

The equation becomes:
10 = 1.645 * (43 / sqrt(n))

To solve for n, we square both sides of the equation and solve for n:

100 = 2.697 * 43^2 / n

n = 2.697 * 43^2 / 100

Calculating this, we get n ≈ 49.4.

Therefore, we need to take a sample size of at least 50 to ensure a 90% probability of getting a sample mean within \$10 of the population mean.