A day 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%

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step-by-step explanation:

hello and welcome!

so, when we see the words "proportional" along with a table, we need to know that we're looking numbers that divide by the same thing.

looking at this table, we simply have to do a trial and error, and divide to see which numbers have the same quotient. if you look at choice a & c, they both have equal quotients for three of the numbers, with one exception, so those can't be the answer. looking at b, there is no similarity, so that can't be it. then, we see d; if you divide all the numbers in the table, you'll see that they all divide by two, which is a proportional relationship, and our answer!

also, always remember that the x will always be the number that goes into y, and somewhere behind this table, is a formula. for this specific table, the formula would be:

since our x-value is being divided by two to get the y-value. and if they were to ask for more numbers in the series, this is the formula you would use to find them.