Question

The distribution of the age, in weeks, at which babies first say "Ma - Ma" is...

The distribution of the age, in weeks, at which babies first say "Ma - Ma" is skewed right with mean 50 weeks and standard deviation 2.5 weeks. A random sample of 45 babies is chosen and the age at which each first crawled is recorded.  Is the sampling distribution of the sample mean normally distributed? why

Homework Answers

Answer #1

Solution :

Given that ,

mean = = 50

standard deviation = = 2.5

n = 45

sample distribution of sample mean is ,

=

= 50

sampling distribution of standard deviation

=  / n = 2.5 / 45=0.37

= 0.37

here sample size greater than 30 mean normal distribution

and population standard dviation give so we apply sampling distribution

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
The following data represent the age​ (in weeks) at which babies first crawl based on a...
The following data represent the age​ (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and s equals = 9.948 weeks. Construct and interpret a 90​% confidence interval for the population standard deviation of the age​ (in weeks) at which babies first crawl. 54,32,43,34,41,27 47,36,56,27,39,28
The data to the right represent the age​ (in weeks) at which babies first crawl based...
The data to the right represent the age​ (in weeks) at which babies first crawl based on a survey of 12 mothers. It is known that the data are normally distributed and that s=13.23 weeks. Construct a 29 59 35 35 51 20 32 54 26 54 44 25 90​% confidence interval for the population standard deviation of the age​ (in weeks) at which babies first crawl. The 90​% confidence interval is (____​ , ____​). ​(Use ascending order. Round to...
The following data represent the age​ (in weeks) at which babies first crawl based on a...
The following data represent the age​ (in weeks) at which babies first crawl based on a survey of 12 mothers. The data are normally distributed and s=9.366 weeks. Construct and interpret a 95​% confidence interval for the population standard deviation of the age​ (in weeks) at which babies first crawl. 49 31 43 34 39 25 46 35 55 26 37 29 Click the icon to view the table of critical values of the​ chi-square distribution. Select the correct choice...
The data shown to the right represent the age​ (in weeks) at which babies first​ crawl,...
The data shown to the right represent the age​ (in weeks) at which babies first​ crawl, based on a survey of 12 mothers. Complete part (b) below. 52 30 44 35 47 37 56 26 35 30 39 (b) Construct and interpret a 95​% confidence interval for the mean age at which a baby first crawls. Select the correct choice and fill in the answer boxes to complete your choice. ​(Round to one decimal place as​ needed.) A.The lower bound...
X is a random variable for the weight of a steak. The distribution of X is...
X is a random variable for the weight of a steak. The distribution of X is assumed to be right skewed with a mean of 3.5 pounds and a standard deviation of 2.32 pounds. 1. Why is this assumption of right skewed reasonable? 2. Why cant you find the probability that a randomly chosen steak weighs over 3 pounds? 3. 5 steaks randomly selected, can u determine the probability that the average weight is over 3 pounds? Explain. 4. 12...
The amount of money collected by a snack bar at a large university has been recorded...
The amount of money collected by a snack bar at a large university has been recorded daily for the past five years. Records indicate that the mean daily amount collected is $3250 and the standard deviation is $400. The distribution is skewed to the right due to several high volume days (including football game days). Suppose that 60 days were randomly selected from the five years and the average amount collected from those days was recorded. Which of the following...
A random variable is not normally distributed, but it is mound shaped. It has a mean...
A random variable is not normally distributed, but it is mound shaped. It has a mean of 17 and a standard deviation of 5. If you take a sample of size 13, can you say what the shape of the sampling distribution for the sample mean is? Why? If the sample size is 13, then you can't say anything about the sampling distribution of the sample mean, since the population of the random variable is not normally distributed and the...
The height of women in the US is normally distributed with mean (mu) = 65 inches...
The height of women in the US is normally distributed with mean (mu) = 65 inches and standard deviation (sigma) = 2.5 inches. A random sample of 15 women is chosen from all women in the US. Is the sampling distribution of the sample ( x - bar) mean normally distributed? Why? A. No because the standard deviation is too small B. No because x < 30 C. Yes. Because x < 30 D. Yes, because the original x distribution...
Suppose students' ages follow a skewed right distribution with a mean of 24 years old and...
Suppose students' ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 5 years. If we randomly sample 250 students, which of the following statements about the sampling distribution of the sample mean age is incorrect? The shape of the sampling distribution is approximately normal. The mean of the sampling distribution is approximately 24 years old. The standard deviation of the sampling distribution is equal to 5 years. All of the above...
A random variable is normally distributed, with a mean of 14 and a standard deviation of...
A random variable is normally distributed, with a mean of 14 and a standard deviation of 3. (a) If you take a sample of size 10, what can you say about the shape of the sampling distribution of the sample mean? Why? (b) For a sample of size 10, state the mean and standard deviation of the sampling distribution of the sample mean. (c) Suppose it turns out that the original distribution (the original random variable) IS NOT EVEN CLOSE...