Question

At school, students spend an average of 13.38 minutes per class texting. Suppose the minutes spent...

At school, students spend an average of 13.38 minutes per class texting. Suppose the minutes spent texting in class are normally distributed with a standard deviation of 3.27 minutes. What is the probability that a randomly selected student spends more than 17 minutes texting in class?

a) 0.8665

b) 0.1335

c) 0.1515

d) 0.8485

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

Homework Answers

Answer #1


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 time, in minutes spent texting in class, is approximately normally distributed with a mean of...
The time, in minutes spent texting in class, is approximately normally distributed with a mean of 6.2 minutes and a standard of 1.6 minutes. The distance, in city blocks, that a student must travel to get from one class to the next has a Chi-squared distribution with 4 degrees of freedom. (The standard city block is approximately 400 feet long) 1) What is the probability that the average time spent texting in a class of 12 random students is between...
17 (CO 3) A survey indicates that shoppers spend an average of 22 minutes with a...
17 (CO 3) A survey indicates that shoppers spend an average of 22 minutes with a standard deviation of 16 minutes in your store and that these times are normally distributed. Find the probability that a randomly selected shopper will spend less than 20 minutes in the store 18(CO 3) The monthly utility bills in a city are normally distributed with a mean of $121 and a standard deviation of $23. Find the probability that a randomly selected utility bill...
Suppose the average time a visitor spends at an Art Gallery is 35 minutes. Assume the...
Suppose the average time a visitor spends at an Art Gallery is 35 minutes. Assume the visit lengths are normally distributed with a standard deviation of 11 minutes. If a visitor is selected at random, what is the probability that they spend: (a) less than 20 minutes at the exhibit? (b) more than 40 minutes at the exhibit?
Suppose college students average 7.3 hours of sleep per night with a standard deviation of 45...
Suppose college students average 7.3 hours of sleep per night with a standard deviation of 45 minutes, and suppose that this is normally-distributed. What is the probability that a randomly-selected college student sleeps for more than 8.1 hours? Probability = What is the 60th percentile for college-student sleep times? Round your answer to four decimal places. 60th percentile = hours
According to Consumer Response Annual Report, millennials spent an average of $103 on monthly dining in...
According to Consumer Response Annual Report, millennials spent an average of $103 on monthly dining in 2016. Let the amount spent on a monthly dining be normally distributed with unknown standard deviation. Assume that the probability of a randomly-selected millennial that spends more than $133 is 10%, and the probability that a randomly-selected millennial that spends less than $97 pounds is 40%. What is the probability that a randomly-selected millennial will spend between $73 and $109 on a monthly dining?
According to Consumer Response Annual Report, millennials spent an average of $103 on monthly dining in...
According to Consumer Response Annual Report, millennials spent an average of $103 on monthly dining in 2016. Let the amount spent on a monthly dining be normally distributed with unknown standard deviation. Assume that the probability of a randomly-selected millennial that spends more than $133 is 10%, and the probability that a randomly-selected millennial that spends less than $97 pounds is 40%. What is the probability that a randomly-selected millennial will spend between $73 and $109 on a monthly dining?
Suppose the amount spent on cell phone service for students per month is normally distributed and...
Suppose the amount spent on cell phone service for students per month is normally distributed and has a mean of $54 and a standard deviation of $9. Binomial or Normal? What is the probability that the monthly cell phone bill for a randomly selected Wake Tech student is more than $60? What is the probability that the monthly cell phone bill for a randomly selected Wake Tech student is less than $50? Most student’s bill will be in the middle...
1. According to an article published on the Web site, Facebook users spend an average of...
1. According to an article published on the Web site, Facebook users spend an average of 190 minutes per month checking and updating their Facebook pages. Suppose that the current distribution of time spent per month checking and updating a member’s Facebook page is normally distributed with a mean of 190 minutes and a standard deviation of 53.4 minutes. For a randomly selected Facebook member, determine the probability that the amount of time that he or she spends per month...
one year consumers spent an average of $22 on a meal at a restaurant. assume that...
one year consumers spent an average of $22 on a meal at a restaurant. assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $3. what is the probability that a randomly selected person spent more than $24. probably that a randomly selected person spends between $5 and $15
the amount of money that students spend on their cell phones per week in normally distributed...
the amount of money that students spend on their cell phones per week in normally distributed with a mean of 52.00 and standard deviation of 6.00. 1.2.1 What is the probability that a student studies for more that 60.00 per week? 1.2.2 find the probability that the mean amount of money on cell phones for three randomly selected students is less than 60.66 per week.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT