Question

A survey recorded the number of children 18 years old or younger who lived in 200...

A survey recorded the number of children 18 years old or younger who lived in 200 households. The accompanying table shows the results.

a. Determine the mean number of children per household.

b. Determine the standard deviation for the number of children per household.

Number_of_Children   Number_of_Families
0 89
1 38
2 41
3 22
4 8
5 2

Homework Answers

Answer #1

Solution-

Let

X = number of children

f = number of families (frequency)

a. mean number of children per household.

= 1.14

b. the standard deviation for the number of children per household.

= 1.2643

Calculations-

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In a survey of 10 households, the number of children was found to be 4, 1,...
In a survey of 10 households, the number of children was found to be 4, 1, 5, 4, 3, 7, 2, 3, 4, 1 (a) State the mode. (b) Calculate (i) The mean number of children per household (ii) The median number of children per household.
A survey was conducted to determine the number of cats living in individual households. The results...
A survey was conducted to determine the number of cats living in individual households. The results are summarized in the accompanying table. Complete parts a and b. Number of Cats Probability 0 0.27 1 0.26 2 0.28 3 0.15 4 0.04 Determine the mean and standard deviation. (Do not round)
A random sample of 4040 adults with no children under the age of 18 years results...
A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.345.34 ​hours, with a standard deviation of 2.372.37 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.124.12 ​hours, with a standard deviation of 1.641.64 hours. Construct and interpret a 9090​% confidence interval for the mean difference in leisure time between adults with no...
A random sample of 45 working mothers with six-year-old or younger children was selected from companies...
A random sample of 45 working mothers with six-year-old or younger children was selected from companies that provide day-care facilities on premises and it was found that they missed an average of 6.4 days from work last year with a standard deviation of 1.2 days. Another sample of 50 such mothers were randomly selected from companies that do not provide day-care facilities on premises and it showed that these mothers missed an average of 9.3 days last year with a...
A survey asked the question​ "What do you think is the ideal number of children for...
A survey asked the question​ "What do you think is the ideal number of children for a family to​ have?" 411 males and 343 females answered the survey and responded with a numeric response from 0 to 6. The 411 males who responded had a mean of 3.18​, and standard deviation of 1.6​, where the 343 females had a median of​ 2, a mean of 3.27 and a standard deviation of 1.69. The​ 95% confidence interval for the female population...
A study of seat belt use involved children who were hospitalized after motor vehicle crashes. For...
A study of seat belt use involved children who were hospitalized after motor vehicle crashes. For a group of 123 children who were wearing seat belts, the number of days in intensive care units (ICU) has a mean of 0.80 and a standard deviation of 1.70. For a group of 290 children who were not wearing seat belts, the number of days spent in ICUs has a mean of 1.30 and a standard deviation of 2.06 (based on data from...
A survey asked a group of medical doctors how many children they had fathered. The results...
A survey asked a group of medical doctors how many children they had fathered. The results are summarized by this ungrouped frequency distribution. Number of Children 0 1 2 3 4 6 Number of Doctors 18 14 24 14 6 3 (a) Calculate the sample mean for the number of children the doctors had fathered. (Give your answer correct to two decimal places.) (b) Calculate the variance for the number of children the doctors had fathered using the short-cut formula...
The number of vehicles passing through a bank drive-up line during each 15 minutes was recorded....
The number of vehicles passing through a bank drive-up line during each 15 minutes was recorded. The results are shown below.                               28   30   28   31 31                               28 33   30   38   34                               34   32 27   34   28                               23   18   30 30   30 a. Compute the sample mean, median, and the mode for the data on distances driven. b.  Determine the range, variance, and standard deviation. c.   Find the first and the third quartiles. State the interquartile range.
According to the American Community Survey, 27% of residents of the United States 25 years old...
According to the American Community Survey, 27% of residents of the United States 25 years old or older had earned a bachelor degree. Suppose you select 12 residents of the United States 25 years old or older and recorded the number who had earned a bachelor degree. Round probabilities to 4 decimal places. Explain why this is a binomial experiment. Find and interpret the probability that exactly 5 of them had a bachelor degree. Find and interpret the probability that...
20 women and 30 men between 18 to 60 years old and the number of hours...
20 women and 30 men between 18 to 60 years old and the number of hours that they work. These information would be your populations. For each group find the followings: 20 Women Ages = 19, 23, 40, 18, 60, 50, 21, 30, 33, 28, 33, 35, 28, 24, 28, 18, 19, 22, 25, 50 20 women work hours weekly = 40, 24, 30, 31, 19, 10, 21, 5, 40, 9, 8, 40, 37, 12, 20, 40, 20, 10, 40,...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT