Question

(CO 1) In a normally distributed data set of how long customers stay in your store,...

(CO 1) In a normally distributed data set of how long customers stay in your store, the mean is 50.3 minutes and the standard deviation is 3.6 minutes. Within what range would you expect 95% of your customers to stay in your store

39.5-61.1

46.7-53.9

43.1-57.5

48.5-52.1

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Given that

mean = 50.3 , s.d = 3.6

Solution :

From empirical rule 95 % of data lies within 2 s.d's from the mean i.e.,

so,

=   50.3 - ( 2 * 3.6 ) = 43.1

50.3 + ( 2 *  3.6 ) = 57.5

Therefore

Answer :  43.1 - 57.5

Add a comment
Know the answer?
Add Answer to:
(CO 1) In a normally distributed data set of how long customers stay in your store,...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • (CO 1) In a normally distributed data set of how long customers stay in your store,...

    (CO 1) In a normally distributed data set of how long customers stay in your store, the mean is 44.8 minutes and the standard deviation is 2.6 minutes. Within what range would you expect 95% of your customers to stay in your store?

  • In a normally distributed data set of how long customers stay in your store, the mean...

    In a normally distributed data set of how long customers stay in your store, the mean is 44.8 minutes and the standard deviation is 2.6 minutes. Within what range would you expect 95% of your customers to stay in your store? a2.6-44.8 b37.0-52.6 c42.2-47.4 d39.6-50.0

  • Please I need help with my assignment 1,In a normally distributed data set of how long...

    Please I need help with my assignment 1,In a normally distributed data set of how long customers stay in your store, the mean is 31.7 minutes and the standard deviation is 1.9minutes . Within what range would you expect 95% of your customers to stay in your store? a.26.0-37.4 b.30.75-32.7 c.29.8-33.6 d.27.9-35.5 2. You need to study the satisfaction of customers of a specific restaurant. You decide to order food and talk to those customers sitting next to you. This...

  • The interarrival times ∆t, for customers into the store are assumed to be all normally distributed...

    The interarrival times ∆t, for customers into the store are assumed to be all normally distributed with mean μ = 8 minutes and standard deviation σ = 2 minutes, so that ∆t ∼ N(μ, σ2) = N(8, 4), in units of minutes. a. Compute Pr(∆t < 0) and use this to argue that it is unnecessary to truncate ∆t so that it is always non-negative. b. Compute the probability that the 16th and the 9th customer arrive within 55 minutes...

  • A. Suppose your manager indicates that for a normally distributed data set you are analyzing, your...

    A. Suppose your manager indicates that for a normally distributed data set you are analyzing, your company wants data points between z=−1.5z=-1.5 and z=1.5z=1.5 standard deviations of the mean (or within 1.5 standard deviations of the mean). What percent of the data points will fall in that range? Answer:___ percent (Enter a number between 0 and 100, not 0 and 1 and round to 2 decimal places) B.  Assume that z-scores are normally distributed with a mean of 0 and a...

  • Suppose your manager indicates that for a normally distributed data set you are analyzing, your company...

    Suppose your manager indicates that for a normally distributed data set you are analyzing, your company wants data points between z=−1.4z=-1.4 and z=1.4z=1.4 standard deviations of the mean (or within 1.4 standard deviations of the mean). What percent of the data points will fall in that range?

  • Suppose your manager indicates that for a normally distributed data set you are analyzing, your company...

    Suppose your manager indicates that for a normally distributed data set you are analyzing, your company wants data points between z=−1.4 and z=1.4 standard deviations of the mean (or within 1.4 standard deviations of the mean). What percent of the data points will fall in that range? Answer:  percent (Enter a number between 0 and 100, not 0 and 1 and round to 2 decimal places)

  • Suppose a normally distributed set of data has a mean of 172 and a standard deviation...

    Suppose a normally distributed set of data has a mean of 172 and a standard deviation of 15. Use the 68-95-99.7 rule to determine the percent of scores in the data set expected to be between the scores 142 and 187. Give your answer in decimal form and keep all decimal places throughout your calculations and in your final answer.

  • Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a...

    Suppose the returns on long-term government bonds are normally distributed. Assume long-term government bonds have a mean return of 6 percent and a standard deviation of 9.6 percent. What is the probability that your return on these bonds will be less than −13.2 percent in a given year? Use the NORMDIST function in Excel® to answer this question. Probability             % What range of returns would you expect to see 95 percent of the time? Expected range of returns            ...

  • 1. According to the empirical rule, in a normally distributed set of data, approximately what percent...

    1. According to the empirical rule, in a normally distributed set of data, approximately what percent of the scores will be within 1 standard deviation (-1 to +1) away from the mean? 40% 95% 68% 75% 2. f you took an IQ test and your score was 2 standard deviations above average, assuming normal distribution, approximately what percent of all IQ test takers would your score be higher than? 98% 60% 70% 80% 3. if you took an IQ test...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT