Question

There is a study that shows that young adults watch a MEAN.P of 6.50 hours of...

There is a study that shows that young adults watch a MEAN.P of 6.50 hours of youtube.com
videos each day with a STDEV.P of 2.50 hours.
Suppose a sample of 36 young adults were chosen, what is the probability (percentage)
that these students watch between 5.50 hours and 5.75 hours of youtube.com videos each day?

Homework Answers

Answer #1

Solution :

= / n = 2.50 / 36 = 0.4167

= P[(5.50 - 6.50) / 0.4167 < ( - ) / < (5.75 - 6.50) / 0.4167)]

= P(-2.40 < Z < -1.80)

= P(Z < -1.80) - P(Z < -2.40)

= 0.0359 - 0.0082

= 0.0277

Probability = 0.0277

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
A survey of 2,250 adults reported that 56% watch news videos. Complete (a) and (b). a)...
A survey of 2,250 adults reported that 56% watch news videos. Complete (a) and (b). a) Suppose that you take a sample of 100 adults. If the population proportion of adults who watch news videos is 0.56, what is the probability that fewer than half in your sample will watch news videos? The probability is ___ that fewer than half of the adults in the sample will watch news videos. b) Suppose that you take a sample of 500 adults....
A survey of 2,350 adults reported that 55% watch news videos. Complete parts​ (a) through​ (c)...
A survey of 2,350 adults reported that 55% watch news videos. Complete parts​ (a) through​ (c) below Suppose that you take a sample of 50 adults. If the population proportion of adults who watch news videos is 0.55 what is the probability that fewer than half in your sample will watch news​ videos? The probability is what? that fewer than half of the adults in the sample will watch news videos The probability is what % that fewer than half...
An imaginary study states that Americans watch an average of 1400 hours of streaming videos each...
An imaginary study states that Americans watch an average of 1400 hours of streaming videos each year. If the standard deviation of the distribution is 221 hours, find the probability that the mean of a randomly selected sample of 50 Americans will be between 1380 and 1430 hours.
In a study on the physical activity of young​ adults, pediatric researchers measured overall physical activity...
In a study on the physical activity of young​ adults, pediatric researchers measured overall physical activity as the total number of registered movements​ (counts) over a period of time and then computed the number of counts per minute​ (cpm) for each subject. The study revealed that the overall physical activity of obese young adults has a mean of μ=320 cpm and a standard deviation of σ=90 cpm. In a random sample of 100 obese young​ adults, consider the sample mean...
A sample of young adults were paused regarding a numbers of hours spent watching Netflix. There...
A sample of young adults were paused regarding a numbers of hours spent watching Netflix. There were 32 adults in the sample with a mean hours wash per week of 11 hours and a standard deviation of one hour. What is the 95% confidence interval for the population mean. a. find both upper and lower bound intervals b. what is the margin of error in the above problem?
Again suppose 20% of OSU students watch reality TV shows of some kind every week. Now...
Again suppose 20% of OSU students watch reality TV shows of some kind every week. Now 100 OSU students are selected at random and asked if they watch reality TV of some kind each week. X = number in the sample who watch reality TV of some kind each week. Find the mean and SD of X. Justify that X has an approximate normal distribution. Use the normal approximation to find the probability that at least 27 of the students...
A random sample of 36 students shows that the mean number of hours spent each night...
A random sample of 36 students shows that the mean number of hours spent each night on homework is 2.5 hours, with a standard deviation of 0.75 hours. (Study time is assumed to be normally distributed.) He has expressed concern that this mean is lower than in the past when 3.0 hours was the norm. Use a one sample test of the means (at 95% confidence) to determine if his concern is justified. Is it?
The day before biology homework is due students come to the study room hours at an...
The day before biology homework is due students come to the study room hours at an average rate of 20 students per hour. Students arrive independently to the study room. a) Let A be the number of students arriving at the study room between 11:30 am and 11:45 am. Find the probability that at least one student comes to the study room during this period. What are the distribution, parameter(s), and support of A? b) Given that at least one...
A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults...
A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. The 95% confidence interval for the mean, μ, is (7.7, 9.3). Which of the following will provide a more informative (i.e., narrower) confidence interval than...
A study report shows that 36% of companies in country A have three or more female...
A study report shows that 36% of companies in country A have three or more female board directors. Suppose you select a random sample of 100 respondents. What is the probability that the sample will have between 32% and 41% of companies in country A that have 3 or more female board directors?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT