Apopulation has a mean of 300 and a standard deviation of 40. suppose a sample of size 100 is selected and is used to estimate . use z-table. what is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)? (round z value in intermediate calculations to 2 decimal places.) what is the probability that the sample mean will be within +/- 11 of the population mean (to 4 decimals)? (round z value in intermediate calculations to 2 decimal places.)

a: 0.9544 9 within 8 units)

b: 0.9940

Step-by-step explanation:

We have µ = 300 and σ = 40.  The sample size, n = 100.  

For the sample to be within 8 units of the population mean, we would have sample values of 292 and 308, so we want to find:

P(292 < x < 308).

We need to find the z-scores that correspond to these values using the given data.  See attached photo 1 for the calculation of these scores.

We have P(292 < x < 308) = 0.9544

Next we want the probability of the sample mean to be within 11 units of the population mean, so we want the values from 289 to 311.  We want to find

P(289 < x < 311)

We need to find the z-scores that correspond to these values.  See photo 2 for the calculation of these scores.

We have P(289 < x < 311) = 0.9940

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

it’s supposed to be -4+45=41

step-by-step explanation: