Question
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 Pizza cost ($) Number of students
10.99-19.99 6
20.99-29.99 20
30.99-39.99 29
40.99-49.99 10
50.99-59.99 10
The sample standard deviation for the cost is $___. (round to the nearest cent)
Homework Answers
Answer #1
| Class | Frequency (f) | Midpoint (xi) | fxi | fxi2 | ||
| 10.99 | - | 19.99 | 6 | 15.49 | 92.94 | 1439.6406 |
| 20.99 | - | 29.99 | 20 | 25.49 | 509.80 | 12994.8020 |
| 30.99 | - | 39.99 | 29 | 35.49 | 1029.21 | 36526.6629 |
| 40.99 | - | 49.99 | 10 | 45.49 | 454.90 | 20693.4010 |
| 50.99 | - | 59.99 | 10 | 55.49 | 554.90 | 30791.4010 |
| Total | 75 | 2641.75 | 102445.9075 | |||
Not the answer you're looking for?
Similar Questions
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 pizza cost $ (Number of students)
10.00-19.99 (9)
20.00-29.99 (13)
30.00-39.99 (28)
40.00-49.99 (12)
50.00-59.99 (5)
The sample standard deviation for the cost is $ ____ (Round to
the nearest cent as needed.)
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 pizza cost ($) Number of students 10.00-19.99 (5)
20.00-29.99 (19) 30.00-39.99 (21) 40.00-49.99 (19)
50.00-59.99
A sample of college students was asked how much they spent monthly on coffee. Approximate the...
A sample of college students was asked how much they spent
monthly on coffee. Approximate the standard deviation for the
cost.
Monthly
coffee
cost ($)
Number of students
10.00-19.99
7
20.00-29.99
20
30.00-39.99
24
40.00-49.99
13
50.00-59.99
6
The sample standard deviation for the cost is __ (Round to the
nearest cent as needed.)
A sample of college students was asked how much they spent monthly on coffee. Approximate the...
A sample of college students was asked how much they spent
monthly on coffee. Approximate the standard deviation for the
cost.
Monthly
coffeecoffee
cost ($)
Number of students
10.00-19.99
7
20.00-29.99
20
30.00-39.99
24
40.00-49.99
13
50.00-59.99
6
The sample standard deviation for the cost is __
A sample of college students was asked how much they spent monthly on cell phone plans....
A sample of college students was asked how much they spent
monthly on cell phone plans.
Approximate the standard deviation for the cost.
Monthly
cell phone plancell phone plan
cost ($)
Number of students
10.00-19.99
8
20.00-29.99
12
30.00-39.99
30
40.00-49.99
12
50.00-59.99
8
The sample standard deviation for the cost is
$______.
(Round to the nearest cent as needed.)
A sample of college students was asked how much they spent monthly on a cell phone...
A sample of college students was asked how much they
spent monthly on a cell phone plan (to the nearest dollar). Find
sample standard deviation.
Monthly Cell Phone Plan Cost ($)
Number of Students
10 - 19
8
20 - 29
16
30 - 39
21
40 - 49
11
50 - 59
4
Q13) The following data represent the difference in
scores between the winning and losing teams in a sample of 15
college football bowl games from 2004-2005....
A marketing professor asked a random sample of college students who have cell phones how much...
A marketing professor asked a random sample of college students
who have cell phones how much their monthly bill was. He found the
sample mean was $100. Using a two-tailed significance test, he
rejected Ho: µ = $80 using the usual decision cutoff of α = 0.05.
Which one of the following could be a 95% confidence interval for
the mean monthly cell phone bill for all college students with cell
phones?
Show your reasoning.
a) ($70, $110)
b) ($70,...
A survey was taken of 383 students that asked the question how much time they spent...
A survey was taken of 383 students that asked the question how
much time they spent watching television in a day (in minutes).
Suppose this sample study follows a Normal distribution with a
sample mean x-bar = 118 minutes. Also suppose the population has a
standard deviation σ = 53 minutes. Determine a 96% confidence
interval for the mean television watching time.
A. 112.67 to 123.31
B. 117.86 to 118.14
C. 112.44 to 123.56
D. 111.70 to 124.30
5000 students that attend local area high schools were asked how much time they spend on...
5000 students that attend local area
high schools were asked how much time they spend on homework each
month. Their answers were recorded in hours, and we’re told that
their answers follow a Normal distribution. The results showed that
the mean time spent on homework each month was 25 hours, with a
standard deviation of 3 hours.
16. What percentage of these students
spent between 25 and 29 hours each month on their homework?
17. What percentage of...
In a study about how much College students sleep at night, a sample of 25 students...
In a study about how much College students sleep at night, a
sample of 25 students was selected. The sample average mean number
of hours of sleep was 6.64 and the sample standard deviation was
1.075.
(a) Construct a 90% confidence interval for the true mean number
of hours of sleep at night College students get.
(b) Interpret the interval.
(c) What assumptions must be satisfied for the interval obtained
in part (a) to be statistically valid?
ADVERTISEMENT
Need Online Homework Help?
Get Answers For Free
Most questions answered within 1 hours.
ADVERTISEMENT
Active Questions
asked 10 minutes ago
asked 10 minutes ago
asked 27 minutes ago
asked 27 minutes ago
asked 29 minutes ago
asked 34 minutes ago
asked 53 minutes ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago
asked 1 hour ago