Question

Suppose the number of customers that show up to a Taco Bell for lunch on a given day is distributed according to a Poisson distribution with lambda =80 customers. Suppose we collect data from 100 days and we calculate the sample mean, xˉ . (a) What is the expected value of the sample mean, E[ Xˉ ] ? (b) What is the variance of the sample mean, Var( xˉ) ? (c) What is the standard error of the sample mean, SE( Xˉ ) ? (d) Using the Central Limit Theorem, what is the (approximate) probability that the sample mean is between at least 79 and at most 81 ?

Answer 1

The expected value of the sample mean,(E[bar{X}] ), is equal to the population mean, which is 80 customers.

Step-by-step explanation:

The variance of the sample mean,(Var(bar{X}) ), is calculated using the formula (Var(bar{X}) = frac{lambda}{n}), where (lambda) is the population mean and (n) is the sample size. Thus, (Var(bar{X}) = frac{80}{100} = 0.8).

The standard error of the sample mean,(SE(bar{X})), is the square root of the variance, so (SE(bar{X}) = sqrt{Var(bar{X})} = sqrt{0.8} approx 0.8944).

Using the Central Limit Theorem, the probability that the sample mean is between at least 79 and at most 81 can be approximated by finding the z-scores for these values and then looking up the corresponding probabilities in a standard normal distribution table. The result is approximately 0.3835.

The expected value of the sample mean ((E[bar{X}])) is equal to the population mean, which is given as (lambda = 80) in the Poisson distribution. Therefore,(E[bar{X}] = lambda = 80).The variance of the sample mean ((Var(bar{X}))) is determined by the formula (Var(bar{X}) = frac{lambda}{n}), where (lambda) is the population mean and (n) is the sample size. Substituting the given values,(Var(bar{X}) = frac{80}{100} = 0.8).

The standard error of the sample mean ((SE(bar{X}))) is the square root of the variance. Therefore,(SE(bar{X}) = sqrt{Var(bar{X})} = sqrt{0.8} approx 0.8944). To find the approximate probability that the sample mean is between at least 79 and at most 81 using the Central Limit Theorem, we calculate the z-scores for these values. Using a standard normal distribution table, the probability is approximately 0.3835.