Question

1. A school administrator sends out grade school students to sell boxes of candy to raise funds. Below is a selection of four students and the mean number of boxes they sold over a weekend. The administrator wants to calculate the average number of boxes sold across students, but wants to weight this by the number of nearby houses (because students with more houses nearby will sell more boxes). For these data, what is the weighted mean?

Mean Candy sold	5	4	18	10
Number of nearby houses	3	4	12	9

2.

Number of songs	Proportion
10	0.1
15	0.14
20	0.15
25	0.11
30	0.13
35	0.16
40	0.09
45	0.07
50	0.05

What is the average expected number of songs from this sample? (the mean of the probability distribution)

3.

Number of songs	Proportion
10	0.1
15	0.14
20	0.15
25	0.11
30	0.13
35	0.16
40	0.09
45	0.07
50	0.05

What is the standard deviation of the number of songs from this sample? (the SD of the probability distribution)

4.

Intervals	Frequency	Cumulative Percent
10-20	1	3
21-30	3	13
31-40	7	35
41-50	10	68
51-60	8	94
61-70	2	100

What number is at the 55th percentile? (You may round to a whole number for the answer)

Answer 1

1) weighted mean =(3*5+4*4+12*18+9*10)/(3+4+12+9)=12.036

2)

Number of songs	Proportion	x*P
10	0.1	1
15	0.14	2.1
20	0.15	3
25	0.11	2.75
30	0.13	3.9
35	0.16	5.6
40	0.09	3.6
45	0.07	3.15
50	0.05	2.5
	total	27.6

expected number of songs =27.6

3)

y	x	P(x)	xP(x)	x2P(x)
1	10	0.1	1.000	10.000
-1	15	0.14	2.100	31.500
-1	20	0.15	3.000	60.000
-1	25	0.11	2.750	68.750
-1	30	0.13	3.900	117.000
-1	35	0.16	5.600	196.000
-1	40	0.09	3.600	144.000
-1	45	0.07	3.150	141.750
-1	50	0.05	2.500	125.000
		total	27.600	894.000
		E(x) =μ=	ΣxP(x) =	27.6000
		E(x2) =	Σx2P(x) =	894.0000
		Var(x)=σ2 =	E(x2)-(E(x))2=	132.240
		std deviation=	σ= √σ2 =	11.4996

standard deviation of the number =11.50

4)

c =class width =							10
n=total frequency =							31
Lpth lower limit of pth percentile interval =							40.5
fcum. Is cumulative frequency till previous interval to pth percentile;							11
fpth is frequency of pth percentile interval							10

55th percentile=

Lpth+c*(np-fcum.)/fpth

=

46.5500~ 47