Question

In: Statistics and Probability

A survey of 200 college students showed the distribution of the number of hours of TV...

A survey of 200 college students showed the distribution of the number of hours of TV watched per week. The mean was found to be 3.66 hours with a standard deviation of 4.93. Assume the normal model is appropriate. Find the number of hours of TV watched by top 25% of college students.

Expert Solution

Given that,

mean = = 3.66

standard deviation = = 4.93

n = 200

= 3.66

= / n =4.93 / 200=0.3486

Using standard normal table,

P(Z > z) = 25%

= 1 - P(Z < z) = 0. 25

= P(Z < z ) = 1 - 0.25

= P(Z < z ) = 0.75

= P(Z <0.67 ) = 0.

z = 0.67 using z table

Using z-score formula

= z * +

= 0.67 *0.3486+3.66

= 3.8936

= 4

orchestra answered 2 years ago

A survey on the number of hours a college student works per week showed that the...

A survey on the number of hours a college student works per week showed that the hours varied from 5 to 61 where 5 was the lowest number and 61 was the highest number. Determine the class width, classes, class marks and class boundaries of a frequency distribution table if the work hours were grouped using 8 classes. You can make a table to help you do this. Note: there is no frequency column.

A survey was conducted at a college to determine the number of hours students spent outside...

A survey was conducted at a college to determine the number of hours students spent outside on weekends during the winter. For students from cold weather climates, the results were M = 8, SD = 1.5. For students from warm-weather climates, the results were M = 5, SD = 4.2. Explain what these numbers mean to a person who has never had a course in statistics, and suggest conclusions the person can draw from these results. These conclusions use the...

A nationwide survey of college students was conducted and found that students spend two hours per...

A nationwide survey of college students was conducted and found that students spend two hours per class hour studying. A professor at your university wants to determine whether the time students spend at your university is significantly different from the two hours. A random sample of 36 statistics students is carried out and the findings indicate an average of 1.75 hours. Assume a populatoin standard deviation of 0.5 hours. Using a level of significance of 0.05: 1. Construct a 95%...

A survey of ten households showed the following number of TV sets per household. 1, 3,...

A survey of ten households showed the following number of TV sets per household. 1, 3, 2, 1, 1, 2, 0, 2, 1, 2 a. Find the mean, variance and standard deviation for this sample of ten households b. Construct a 95% confidence interval for the population mean number of TV sets per household. c. Test if the population mean number of TV sets per household is 2.

10. A survey of 145 randomly selected students at one college showed that only 87 checked...

10. A survey of 145 randomly selected students at one college showed that only 87 checked their campus email account on a regular basis. Construct and interpret a 90% confidence interval for the percentage of students at that college who do not check their email account on a regular basis. (1pt) Point Estimate: ______ Margin of Error: ________ Confidence Interval:_________________ Interpretation: ____

Fifteen randomly selected college students were asked to state the number of hours they slept the...

Fifteen randomly selected college students were asked to state the number of hours they slept the previous night. The resulting data are 5, 10, 7, 8, 6, 7, 7, 6, 6, 9, 4, 3, 8, 9, 7. Find the following. (Round your answers to two decimal places.) (a) Variance s2, using formula below $ s^2 = \dfrac{\sum{(x - \overline{x})^2}}{n - 1}$ s2 = (b) Variance s2, using formula $ s^2 = \dfrac{\sum{x^2} - \dfrac{(\sum{x})^2}{n}}{n - 1} $ s2 = (c)...

3. A local college is studying the number of credit hours students take and how many...

3. A local college is studying the number of credit hours students take and how many total minutes they spend reading for fun/pleasure (e.g., non-class-related materials). They don’t know whether to expect that studentswill read more or less when enrolled in a lot of classes/hours or fewer hours. They study 68 students and find a correlation of -0.30. (A) State the decision rule for the 0.05 significance level. (6) Reject H0 if t > _________ (B) Compute the value of...

The age distribution of students at a community college is given below. Age (years) Number of...

The age distribution of students at a community college is given below. Age (years) Number of Students (f) Under 21 416 21-25 420 26-30 219 31-35 51 Over 35 24 Total = 1130 Number of students (f) 416 420 219 51 24 1130 A student from the community college is selected at random. Find the conditional probability that the student is at most 35 given that he or she is at least 26.

A class survey in a large class for first‑year college students asked, “About how many hours...

A class survey in a large class for first‑year college students asked, “About how many hours do you study during a typical week?” The mean response of the 463 students was ?¯=13.7 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation ?=7.4 hours in the population of all first‑year students at this university. Regard these students as an SRS from the population of all first‑year students at this university. Does the study give...

A class survey in a large class for first-year college students asked, "About how many hours...

A class survey in a large class for first-year college students asked, "About how many hours do you study in a typical week?". The mean response of the 427 students was x¯¯¯ = 12 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation 7 hours in the population of all first-year students at this university. What is the 99% confidence interval (±0.001) for the population mean? Confidence interval is from ____to_____ hours.