Question

Proctor & Gamble reported that an American family of four washes an average of 1 ton...

Proctor & Gamble reported that an American family of four washes an average of 1 ton (2000) pounds of clothes each year. If the standard deviation of the distribution is 187.5 pounds, find the probability that the mean of a randomly selected sample of 50 families of four will be between 1980 and 1995 pounds. Round final answer to 4 decimal places (0.0001).

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Solution :

Given that ,

mean = = 2000

standard deviation = = 187.5

= / n = 187.5 / 50 = 26.5165

= P[(1980 - 2000) /26.5165 < ( - ) / < (1995 - 2000) / 26.5165)]

= P(-0.75 < Z < -0.19)

= P(Z < -0.19) - P(Z < -0.75)

= 0.4247 - 0.2266

= 0.1981

Probability = 0.1981

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Procter & Gamble reported that an American family of four washes an average of 2100 pounds...
Procter & Gamble reported that an American family of four washes an average of 2100 pounds of clothes each year. If the standard deviation of the distribution is 187.5 pounds, find the probability that the mean of a randomly selected sample of 50 families will be a. between 2020 and 2150 pounds b. less than 2130 pounds c. greater than 2050 pounds
A survey found that the American family generates an average of 17.2 pounds of glass garbage...
A survey found that the American family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 45 families will be between 16 and 17 pounds. Assume that the sample is taken from a large population and the correction factor can be ignored. Use a TI-83 Plus/TI-84 Plus calculator and round the final answers to at least four decimal...
a survey found that the american family generates an average of 17.2 pounds of glass garbage...
a survey found that the american family generates an average of 17.2 pounds of glass garbage each year. Assume the standard deviation of the distribution is 2.5 pounds. Find the probability that the mean of a sample of 30 families will be between 16.4 and 17.4 pounds. The correction factor can be ignored
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $69.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). ( , )...
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $75.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). (  ,  ) b. Based...
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $78.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). ( ___ ,...
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $74.50. a. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). (____,_____) b. Based...
According to a survey conducted by the American Automobile Association, a family of four spends an...
According to a survey conducted by the American Automobile Association, a family of four spends an average of $255.60 per day while on vacation in the US. Suppose a sample of 36 families of four vacationing at Mount Rushmore resulted in a sample mean of $235.31 per day and a sample standard deviation of $53.50. a) Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Mount Rushmore (round to two...
1) American ladybugs have an average adult length of 1 cm with a known standard deviation...
1) American ladybugs have an average adult length of 1 cm with a known standard deviation of 0.2 cm. The population of American ladybugs in Raleigh was around 2176200 last spring. Assume a normal distribution for the lengths of adult American ladybugs. Your niece asks you what's the probability of a random ladybug in Raleigh being bigger than 1.5 cm. Is it appropriate to calculate this probability? Select one: a. No, because the sample size is less than 30. b....
1 . A study of 5,000 American women was taken to determine the mean number of...
1 . A study of 5,000 American women was taken to determine the mean number of children per household. Samples of 100 were taken by state. The average American household, as reported by women, contains 2.1 children with a standard deviation of 1.87. What would be the mean of the sampling distribution of the sample means? (If rounding is necessary, round to four decimal places). 2 . A study of 5,000 American women was taken to determine the mean number...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT