Question

The average time a visitor spends at the Renzie Park Art Exhibit is 62 minutes

The average time a visitor spends at the Renzie Park Art Exhibit is 62 minutes. The standard deviation is 12 minutes. If a visitor is selected at random, find the probability that he or she will spend:

A)At least 82 minutes at the exhibit
B)At most 50 minutes at the exhibit




The average time a person spends at Barefoot Landing Seaquarium is 96 minutes. The standard deviation is 17 minutes. Assume the variable is normally distributed. If a visitor is selected at random, find the probability that he or she will spend:

A)At least 120 minutes at the exhibit
B)At most 80 minutes at the exhibit
1 1
Add a comment Improve this question Transcribed image text
✔ Recommended Answer
Answer #1

X - Average time a visitor spends at the Renzie Park Art Exhibit is N(62, 12)

a) P(X>=82) = P(Z>=1.67)

=0.0475

= 4.75%

-----------------------------------------------------

b) P(X<50) = P(Z<-1) =

0.1587

=15.87%

Add a comment
Know the answer?
Add Answer to:
The average time a visitor spends at the Renzie Park Art Exhibit is 62 minutes
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Similar Homework Help Questions
  • 1.  A national study finds that a US adult spends an average of 48 minutes a day...

    1.  A national study finds that a US adult spends an average of 48 minutes a day on the phone with a standard deviation of 9 minutes.  This data is normally distributed. a)  Find the probability that a randomly selected US adult spends more than 60 minutes on the phone a day. b)  Find the probability that a randomly selected US adult spends between 32 and 40 minutes on the phone each day. Zina includes her answer below.  What went wrong? P(z < -.89) =...

  • People from Missouri spends an average of 49 minutes with an STD (Standard Deviation) of 12...

    People from Missouri spends an average of 49 minutes with an STD (Standard Deviation) of 12 minutes in a grocery store (normally distributed). Find the probability that the shopper will be in the store for more than 52 minutes? (Draw the normal curve with labels and shade the appropriate area)

  • The amount of time that a drive-through bank teller spends on a customer is a random...

    The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean 3.2 minutes and a standard deviation a = 1.6 minutes. If a random sample of 64 customers is observed, find the probability that their mean time at the teller's window is A. at most 2.7 minutes; B. more than 3.5 minutes; C. at least 3.2 minutes but less than 3.4 minutes (10 pts. each, 30 pts. total)

  • Suppose a geyser has a mean time between eruptions of 62 minutes. Let the interval of...

    Suppose a geyser has a mean time between eruptions of 62 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 24 minutes. Complete parts (a) through (e) below. (a) What is the probability that a randomly selected time interval between eruptions is longer than 72 minutes? The probability that a randomly selected time interval is longer than 72 minutes is approximately 0.3372. (Round to four decimal places as needed.) (b) What is the probability...

  • The average employee in the U.S.A. spends μ = 23 minutes commuting to work each day....

    The average employee in the U.S.A. spends μ = 23 minutes commuting to work each day. Assume that the distribution of commute times is normal with a standard deviation of σ = 8 minutes. (a) What proportion of U.S. employees spend less than 15 minutes a day commuting? (b) What is the probability of randomly selecting an employee who spends more than 35 minutes commuting each day?

  • The amount of time that a drive-through bank teller spends on a customer is a random...

    The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean u = 7.9 minutes and a standard deviation o = 3.6 minutes. If a random sample of 81 customers is observed, find the probability that their mean time at the teller's window is (a) at most 7.3 minutes; (b) more than 8.7 minutes; (c) at least 7.9 minutes but less than 8.3 minutes. Click here to view page 1 of...

  • the Idea in details Thank you! The amount of time (in minutes) that Dr. Swift spends...

    the Idea in details Thank you! The amount of time (in minutes) that Dr. Swift spends helping each student that visits during office hours is a random variable having an exponential distribution with 0-11. The (a) When he arrives for office hours on Monday, there is 1 student waiting for help. What (b) When he arrives for office hours on Tuesday there are 2 students waiting for help time helping each student is independent of the time helping any other...

  • A survey indicates that for each trip to the supermarket, a shopper spends an average of...

    A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for each interval of time below: Between 24 and 54 minutes More than 39 minutes

  • 13 . According to the American Time Use Survey, the typical American spends 154.8 minutes per...

    13 . According to the American Time Use Survey, the typical American spends 154.8 minutes per day watching television. A survey of 50 Internet users results in a mean time watching television per day of 128.7 minutes, with a standard deviation of 46.5 minutes. Using 0.05 level of significance to test the claim that Internet users spend less time watching television. 14. College math instructors suggest that students spend 2 hours outside class studying for every hour in class. So,...

  • 2. The amount of time (in minutes) that Dr. Swift spends helping each student that visits...

    2. The amount of time (in minutes) that Dr. Swift spends helping each student that visits during office hours is a random variable having an exponential distribution with 0 11. The (a) When he arrives for office hours on Monday, there is 1 student waiting for help. What (b) Whe he arrives for office hours on Tuesday there are 2 students waiting for help time helping each student is independent of the time helping any other student is the probability...

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