Aday care program has an average daily expense of $75.00. the standard deviation is $15.00. the owner takes a sample of 64 bills. what is the probability the mean of his sample will be between $70.00 and $80.00?
step 1. calculate a z-score for $70.00:
step 2. give the probability for step 1:
step 3. calculate the z-score for $80.00:
step 4. give the probability for step 3:
step 5. add the probabilities from steps 1& 3:

The sample mean is approximately normally distributed with mean $75.00 and standard deviation $15.00/sqrt(64)

1) The z-score for $70.00 is

(70 - 75)/(15/8) = -2.66667 = -2.7 to the nearest tenth.

2) The probability corresponding to this z-score is 0.0035 (use standard normal tables), or 0.35%, which rounded to tenths is 0.4%

I will leave the rest to you. The instruction in step 5 is wrong; you do not add the probabilities from steps 1 and 3. (If you do add them, you will get 1, which is clearly the wrong answer.)

P(70 < sample mean < 80) = P(sample mean < 80) - P(sample mean < 70)

so you should subtract the result in step 1 from the result in step 3.

The correct answer is 0.993, or 99.3%

Ibelieve it is