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  3. 1. a school administrator sends out grade school s

1. A school administrator sends out grade school students to sell boxes of candy to raise...

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Statistics

1. A school administrator sends out grade school students tosell boxes of candy to raise funds. Below is a selection of fourstudents and the mean number of boxes they sold over a weekend. Theadministrator wants to calculate the average number of boxes soldacross students, but wants to weight this by the number of nearbyhouses (because students with more houses nearby will sell moreboxes). 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 thissample? (the SD of the probability distribution)

4.

IntervalsFrequencyCumulative Percent
10-2013
21-30313
31-40735
41-501068
51-60894
61-702100

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

Answer & Explanation Solved by verified expert
4.0 Ratings (435 Votes)

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
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