The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.6 years and a standard deviation of 1.9 years. Random samples of size 18 are drawn from the population and the mean of each sample is determined. Find the mean and the standard error.

a) 5.6 years

b) 0.4478

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 5.6 years

Standard Deviation, σ = 1.9 years

Sample size, n = 18

We are given that the distribution of amounts of time is a bell shaped distribution that is a normal distribution.

a) Mean of each sample.

The best estimator of the sample mean is the population mean itself. Thus, we have:

Thus, the sample mean is 5.6 years.

b) Standard error

Formula:

Putting values, we have.

Thus, the standard error is 0.4478.

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!