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A day care program has an average daily expense of $75.00. The standard deviation is $5.00. 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 2&4. %

Answer :

Answer:

What is  the probability the mean of his sample will be between $70.00 and $80.00? = 64/64 =1

s1) = 99% =z score = 127/128

s2)64/64 = 1

s3)95% of 64 = 123/128

s4) p) step 3 = 95% of 64 = 60.08 /64 = 60/64 = 93.75%

s5) P)of 2+4 = 283/292 = 96.92%

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