The manager of a grocery store has taken a random sample of 100 customers.The average length of time it took these 100 customers to check out was 3 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.With a .95 probability, the sample mean will provide a margin of error of
A) 1.96.
B) .10.
C) .196.
D) 1.64.

Question

Quizplus · Accepted Answer

We can use the formula for margin of error:

Margin of Error = z* (standard deviation / sqrt(sample size))

We are given the sample size (n = 100) and the standard deviation (σ = 1 minute), but we need to find the value of z for a .95 probability.

The z-value for a .95 probability (two-tailed) is 1.96.

So, plugging in the values:

Margin of Error = 1.96 * (1 / sqrt(100))

Margin of Error = 0.196

Therefore, the margin of error is 0.196, which corresponds to answer choice C.

The Manager of a Grocery Store Has Taken a Random