The stem in a stem-leaf diagram

The trunk in the trunk diagram gives the right digit of the number you want and the leaf gives the right digit of the number you want. To achieve the appropriate value, the leaf unit must be multiplied by each leaf value. In our example, the lowest value is given by the least value of stem and leaf, which are 5 and 2. So, the smallest value is 5(2 × 0.1) = 5.2. Proceeding similar way, the largest number may be found from the largest (lowermost) value of stem and the largest value of the leaf. This gives 8 and 9 and the largest number becomes 8(9 × 0.1) = 8.9.
Problem 6
We have 28 numbers (as can be found from the leaf values). So, each quartile will be composed of 7 values. The first quartile be given by first 7 values constructed from the stem and leaf.
The average of 7th and 8th values will give the first quartile value. Thus, the average of 6.1 and 6.4 is 6.25, which is the required answer.
Problem 9
Sample mean may be found by adding all numbers and then dividing the by the number of terms. Here,
21 + 38 + 35 + 31 + 28 + 22 + 25 + 32 + 24 + 35 + 27+ 30 = 348
Total number of terms = 12
Average = 348/12 = 29.0
Problem 10
Sample median may be found by arranging the terms in ascending order and finding the value of the middle term.
21, 22, 24, 25, 27, 28, 30, 31, 32, 35, 35, 38
Here, total number of terms is 12; middle terms are 6th and 7th terms: 28 and 30. The average of the middle terms = (28+30)/2 = 58/2 = 29.0
Problem 11
Sample standard deviation may be found by
Subtracting the average value of the sample from each term
Taking the sum of squares of all differences,
Dividing the sum by the number of terms less one (here, 12-1 =11)
Then taking square root of the result.
Standard Deviation = √(((21–29)^2+(22–29)^2+(24–29)^2+(25–29)^2+(27–29)^2+(28–29)^2+(30–29)^2+(31–29)^2+(32–29)^2+(35–29)^2+(35–29)^2+(38–29)^2)/11)
= 5.44
Problem 12
Coefficient of variation is found by taking percentage of the quotient of standard deviation by mean.
Coefficient of variation = (Standard Deviation )/(Mean ) × 100%
= 5.44/29 × 100%
= 18.8%
Problem 16
z-score = (x–μ)/σ; here, mean, μ = 29 and standard deviation, σ = 5.44
= (38–29)/5.44
= 1.65
Problem 20
Slope, b1 = (n∑▒tY–(∑▒〖Y)(∑▒〖t)〗〗)/(n∑▒t^2 –(∑▒〖t)〗^2 )
Here, ∑▒t = 1 + 2 + 3 + 4 +5 = 15
∑▒t^2 = 12 +22 +32 +42 +52 = 1 +4 +9 +16 +25 = 55
= (5×1104.2–335.7×15)/(5×55 –〖15〗^2 ) = 9.71
Problem 21
Intercept, b0 = (∑▒Y)/n – b1 (∑▒t)/n
= (1104.2)/5 – 9.71×15/5
= 38.01
Problem 22
The trend equation, Y’ = b0 + b1t
= 38.01 + 9.71t
For 2006, t = 6; the forecasted number of visitor can be found by putting t = 6 in the trend equation.
Y’ = 38.01 + 9.71 × 6 = 96.03
PRACTICE EXAM 1-B
Problem 6
The stem in a stem-leaf diagram gives the rightmost digit of the desired number and the leaf will give the leftmost digit of the desired number. The leaf unit must be multiplied with each leaf value to get the required value.
In our example, the lowest value is given by the least value of stem and leaf, which are 2 and 2. So, the smallest value is 2(2 × 10) = 220. Total number of leaves before 4 in the stem row 4 are 17.
Problem 10
Sample mean may be found by adding all numbers and then dividing the by the number of terms. Here,
39 + 13 + 20 + 5 + 29 + 3 + 7 + 11 + 8 + 9 + 35 + 17 + 5 + 40 + 44 = 285
Total number of terms = 15
Average = 285/15 = 19.0
Problem 11
Sample median may be found by arranging the terms in ascending order and finding the value of the middle term.
3, 5, 5, 7, 8, 9, 11, 13, 17, 20, 29, 35, 39, 40, 44
The middle term of 15 terms is the 8th term, which is 13.
Problem 12
See problem 11 of exam 1-A
Problem 16
See problem 16 of exam 1-A
Problem 19
Problem 20
See problem 21 of exam 1-A
Problem 21
See problem 22 of exam 1-A
Problem 22
Running total is calculated by adding all quarter values corresponding to each year.
The first running total corresponds to first year 2003:
175 + 167 + 140 + 135 = 617
The first running total corresponds to first year 2004:
180 + 175 + 150 + 142 = 647
PRACTICE EXAM 1-C
Problem 8
See problem 5 of exam 1-A
Problem 9
The values are 410 to 960, which gives a range of (960–410) = 550.
Problem 10
See problem 10 of exam 1-B
Problem 11
See problem 11 of exam 1-B
Problem 12
See problem 12 of exam 1-B
Problem 15
See problem 16 of exam 1-B
Problem 19
See problem 20 of exam 1-A
Problem 20
See problem 21 of exam 1-A
Problem 21
See problem 22 of exam 1-A
PRACTICE EXAM 1-D
Problem 4
See problem 5 of exam 1-A
Problem 8
See problem 9 of exam 1-A
Problem 9
See problem 10 of exam 1-A
Problem 10
See problem 11 of exam 1-A
Problem 11
See problem 12 of exam 1-A
Problem 13
See problem 16 of exam 1-A
Problem 17
See problem 20 of exam 1-A
Problem 18
See problem 21 of exam 1-A
Problem 19
See problem 22 of exam 1-A
PRACTICE EXAM 1-E
Problem 6
See problem 5 of exam 1-A
Problem 9
See problem 9 of exam 1-A
Problem 10
See problem 10 of exam 1-A

Problem 11

See problem 11 of exam 1-A

Problem 12

See problem 12 of exam 1-A

Problem 15

See problem 16 of exam 1-A

Problem 19

See problem 20 of exam 1-A

Problem 20

See problem 21 of exam 1-A

Problem 21

See problem 22 of exam 1-A