Question

3. Suppose a random sample of text-message lengths (in characters) resulted in the following observations: 50,44,43,59,41....

3. Suppose a random sample of text-message lengths (in characters) resulted in the following observations: 50,44,43,59,41. What was this sample's standard deviation?

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

Homework Answers

Answer #1

3)

Solution :

n = 5

= = (50 + 44 + 43 + 59 + 41) / 5 = 47.4

s = (x - )2 / n - 1  

=(50 - 47.4)2 + (44 - 47.4)2 + (43 - 47.4)2+ (59 - 47.4)2​​​​​​​+ (41 - 47.4)2​​​​​​​/ (5 - 1)

= 53.3

= 7.300

Sample's standard deviation = 7.300

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
The lengths of text messages are normally distributed with a population standard deviation of 3 characters...
The lengths of text messages are normally distributed with a population standard deviation of 3 characters and an unknown population mean. If a random sample of 26 text messages is taken and results in a sample mean of 29 characters, find a 98% confidence interval for the population mean. Round your answers to two decimal places. z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576 You may use a calculator or the common z-values above. Select the...
The lengths of text messages are normally distributed with a population standard deviation of 3 characters...
The lengths of text messages are normally distributed with a population standard deviation of 3 characters and an unknown population mean. If a random sample of 29 text messages is taken and results in a sample mean of 30characters, find a 92% confidence interval for the population mean. Round your answers to two decimal places z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576 select the correct answer below: (28.56,31.44) (29.29,30.71) (29.08,30.92) (28.70,31.30) (29.02,30.98) (28.91,31.09)
The lengths of text messages are normally distributed with a population standard deviation of 4 characters...
The lengths of text messages are normally distributed with a population standard deviation of 4 characters and an unknown population mean. If a random sample of 27 text messages is taken and results in a sample mean of 23 characters, find a 95% confidence interval for the population mean. Round your answers to two decimal places. z0.10 z0.05 z0.04 z0.025 z0.01 z0.005 1.282 1.645 1.751 1.960 2.326 2.576
Suppose lengths of text messages have an unknown distribution with mean 30 and standard deviation 4...
Suppose lengths of text messages have an unknown distribution with mean 30 and standard deviation 4 characters. A sample of size n=61 is randomly taken from the population and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?
A simple random sample of 30 items resulted in a sample mean of 25. The population...
A simple random sample of 30 items resulted in a sample mean of 25. The population standard deviation is 3. Round your answers to two decimal places. a. What is the standard error of the mean, ? b. At 95% confidence, what is the margin of error?
Suppose a random sample of n = 25 observations is selected from a population that is...
Suppose a random sample of n = 25 observations is selected from a population that is normally distributed with mean equal to 108 and standard deviation equal to 14. (a) Give the mean and the standard deviation of the sampling distribution of the sample mean x. mean     standard deviation     (b) Find the probability that x exceeds 113. (Round your answer to four decimal places.) (c) Find the probability that the sample mean deviates from the population mean ? = 108...
Suppose a random sample of n = 16 observations is selected from a population that is...
Suppose a random sample of n = 16 observations is selected from a population that is normally distributed with mean equal to 102 and standard deviation equal to 10. a) Give the mean and the standard deviation of the sampling distribution of the sample mean x. mean = standard deviation = b) Find the probability that x exceeds 106. (Round your answer to four decimal places.) c) Find the probability that the sample mean deviates from the population mean μ...
A simple random sample of 40 items resulted in a sample mean of 10. The population...
A simple random sample of 40 items resulted in a sample mean of 10. The population standard deviation is = 6. Round your answers to two decimal places. a. What is the standard error of the mean, ? b. At 95% confidence, what is the margin of error?
Suppose that a time study analyst takes a sample of size 60 observations. The process mean...
Suppose that a time study analyst takes a sample of size 60 observations. The process mean is 100 seconds, and the standard deviation is 3 seconds. If the max error can be no more than ± 1 second, how confident can the analyst be in the results? *text,no pic, ty
A simple random sample of 10 items resulted in a sample mean of 20. The population...
A simple random sample of 10 items resulted in a sample mean of 20. The population standard deviation is σ = 6. Round your answers to two decimal places. a. What is the standard error of the mean, σx? b. At 95% confidence, what is the margin of error?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT