2.) For the following table, what is the value of :
a) P(A1) = 3/10 = 0.3
b) P(B1│A2) = 1/3
c) P(B2 and A3). Compute this as P(B2)*P(A3│ B2) . Rows are B1 & B2: columns are A1, A2
& A3.
What are the values of P(B2) , P(A3|B2) and P(B2&A3)?
In what row and column will you find this answer?
Second Event
A1
B1
B2
Total
2
1
3
First Event
A2
1
2
3
A3
Total
3
1
4
6
4
10
P(B2) = 4/10 = 0.4
P(A3|B2) = ¼ = 0.25
P(B2 and A3) = 0.4*0.25 = 0.1
3. The chart below gives the percentage of counties in the US that use various methods for
recording votes in 1980 & 2002. Set up side-by-side bar charts by year
METHOD
1980
2002
Punch cards
18.5
15.5
Lever machines
36.7
10.6
Paper ballots
40.7
10.5
Optical scan
0.8
43.0
Electronic
0.2
16.3
Mixed
3.1
4.1
The required chart is given below,
SIde by Side Bar chart for years
50
45
40
35
30
25
20
15
10
5
0
Punch
cards
Lever
machines
Paper
ballots
Optical
scan
Electronic
Mixed
1980
18.5
36.7
40.7
0.8
0.2
3.1
2002
15.5
10.6
10.5
43
16.3
4.1
4. The following are a sample of the weights of nine jars of peanut butter.
7.69 7.72 7.80 7.86 7.90 7.94 7.97 8.06 8.09
a) Compute the median weight.
The median weight is 7.9.
b) Compute the standard deviation of the sample using the shortcut formula. Show the
formula and values for each term and compute the answer. .
1
Standard deviation = √𝑛 ∑ 𝑉𝑎𝑙𝑢𝑒 2 − 𝑀𝑒𝑎𝑛2
Now mean = 7.892
So,
1
Standard deviation = √9 (7.692 + ⋯ + 8.092 ) − 7.8922 = 0.13155
c) Compute the 5 Number System for this data. Just list the values as Xmin=xxx,
Q1=xxx,Q2=xxxx,Q3=xxxxx, Xmax = xxxx.
The 5 Number summary is given below,
Xmin
Q1
Q2
Q3
Xmax
7.69
7.8
7.9
7.97
8.09
d) Are there outliers? An outlier value is defined as unusually large or small according to the
expressions: Answer yes or no.
Outlier > Q3 + 1.5(Q3-Q1) or,
Outlier < Q1-1.5(Q3-Q1)
No there are no outliers
5. Answer questions a. through f. of 3.23 below.
Show work for part f. only.
6. Compute the mean and standard deviation for the data in the table below. State your
assumptions and show all calculations.
Distance
Frequency
0 to 5
5 to 10
10 to 15
15 to 20
20 to 25
4
15
27
18
6
As it is grouped data so to calculate the mean and standard deviation we need to calculate the
class mid points. I am calculating the mid points using the formula,
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓 𝑡ℎ𝑎𝑡 𝑐𝑙𝑎𝑠𝑠+𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑜𝑓 𝑡ℎ𝑎𝑡 𝑐𝑙𝑎𝑠𝑠
Mid Point of a class =
2
Using this formula the required values are given below,
Distance
0 to 5
5 to 10
10 to 15
15 to 20
20 to 25
Total
Frequency Mid Point
(fi)
(xi)
Xifi
xi^2*fi
4
2.5
10
25
15
7.5
112.5
843.75
27
12.5
337.5 4218.75
18
17.5
315
5512.5
6
22.5
135
3037.5
70
-
910
13637.5
Now,
Mean = Total(xifi)/Total(fi) = 910/70 = 13
Variance = Total(xi^2fi)/Total(fi) – Mean^2 = 13637.5/70 – 13^225.82143
SD = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 5.08148
7. A box contains 3 red balls and 4 green balls. If two balls are randomly selected in sequence,
what is the possibility that a red ball and a green ball are picked out of the box. State the rule for
P(A&B) and then substitute the values and compute the answer.