Class interval | Midpoint, x | Frequency, f | fx | fx² |
10 - 19.99 | 14.995 | 10 | 149.95 | 2248.50025 |
20 - 29.99 | 24.995 | 18 | 449.91 | 11245.50045 |
30 - 39.99 | 34.995 | 26 | 909.87 | 31840.90065 |
40 - 49.99 | 44.995 | 11 | 494.945 | 22270.05028 |
50 - 59.99 | 54.995 | 8 | 439.96 | 24195.6002 |
Total | 73 | 2444.635 | 91800.5518 |
Standard deviation, s = √((∑fx² - (∑fx)²/n)/(n-1))
= √((91800.5518 - 2444.635²/73)/(73- 1)) =
11.75
A sample of college students was asked how much they spent monthly on pizza. Approximate the...
A sample of college students was asked how much they spent monthly on pizza. Approximate the standard deviation for the cost. Monthly pizzapizza cost ($) Number of students 10.00-19.99 9 20.00-29.99 14 30.00-39.99 26 40.00-49.99 15 50.00-59.99 5 The sample standard deviation for the cost is (Round to the nearest cent as needed.)
Monthly coffeecoffee cost ($) Number of students 10.00-19.99 10 20.00-29.99 18 30.00-39.99 27 40.00-49.99 12 50.00-59.99 7 A sample of college students was asked how much they spent monthly on coffee. Approximate the standard deviation for the cost. The sample standard deviation for the cost is..... (Round to the nearest cent as needed.)
Objective 3: Approximate the Standard Deviation from a Frequency Distribution O of 1 Point 3.3.RA-3 Question Help A sample of college students was asked how much they spent monthly on cell phone plans. Approximate the standard deviation for the cost. Monthly cell phone plan cost ($) Number of students 10.00-19.99 8 20.00-29.99 15 30.00-39.99 22 40.00-49.99 10 50.00-59.99 9 (Round to the nearest The sample standard deviation for the cost is $ cent as needed.) Q
7. A sample of college students were asked how much pizza they bought in a week. The results were put in a probability distribution table. a) Find P (x = 4). Find P (x < 2). b) Find the expected value of the number of pizzas in a week. Interpret the answer. x P (x) 0 0.10 1 0.34 2 0.23 3 0.15 4 0.18
A sample of 200 college students are asked about how much time they spend studying for classes each week. The results show that the sample has a mean of 15 hours with a standard deviation of 5 hours. Explain these findings for someone who has never taken a statistics class before. Note that your job is to interpret the findings, NOT to explain how they were calculated! (10 points)
number one has been answered! please help with the rest
2- A randon sample of College students were asked how moch money they spent on books and supplies this senester. Beze are the sumary statistics: a-10 $249, s-$30.construct a 90% confidence interval tor the nean coats for all college students. Assune that the anount spent on books and supplies by college students is nonally distributed. 3- A researchez wants to estinate the mean beights of 8- year- old gizis. It...
Random sample of college students were asked how many hours they spent studying per day. The results summarized by sex are given in the following table. Do these data provide evidence at the 0.01 level of significance to conclude that sex and amount of time spent studying are dependent? Begin by filling in the following tables.
1. To determine the average number of hours spent studying by college students per week, a sample of 39 students was randomly selected, and found to spend an average of 17.1 hours per week, with a standard deviation of 4.3 hours. Find the 90% confidence interval for the mean number of hours spent studying per week by all college students. What is the upper and lower bound? 2. If I asked a random student how many hours they study per...
A random sample of community college students was asked the number of hours they sleep on a typical week-night during a given academic term. The sample data are as follows: 8 6 4 5 3 7 S 4 3 4 4 5 6 8 7 7 7 3 3 4 What is the 90% confidence interval estimate for the true mean amount of sleep time per night spent by community college students during a academic term? a) The data give...
3. How much TV do College Students watch? In a survey 361 students were asked how many hours of TV they watched per week. The sample gave an average of 6.504 hours. A bootstrap distribution gave a SE of 0.2939. Find and interpret a 90% confidence interval for the average number of hours a student watches TV per week.