Posted: August 26th, 2021
Student’s Name
Instructors Name
Course
Date
Math
Worksheet for Chapter 12
Part A
Question 1. Construct a frequency distribution table
CLASS | FREQUENCY |
1 – 8 | 0 |
9-12 | 10 |
13-15 | 6 |
16-19 | 9 |
20-23 | 5 |
Question 2
STEM | LEAF |
0 | 9 |
1 | 0,4,5,0,9,5,6,7,3,3,2,0,4,6,7,8,9,0,7,9,1,2 |
2 | 0,1,0,1,1 |
Question 3
a) Calculate the mean
Mean=μ = (ΣXi) / N
=447/30
=14.9
b) Calculate the median
9 ,9 ,9 ,10 ,10 ,10 ,10 ,11 ,12 ,12 ,13 ,13 ,14 ,14 ,15 ,15 ,16 ,16 ,17 ,17 ,17 ,18 ,19 ,19 ,19 ,20 ,20 ,21 ,21 ,21
Median=14+15/2
=14.5
c) Calculate the mode
The mode is the number with the most number of appearances. In this data, 10 appears 10 times. Thus, the mode is 10
d) Calculate the Standard Deviation
σ = sqrt [Σ(Xi – μ) 2 / N]
μ | Xi – μ | (Xi – μ )2 |
14.9 | 10-14.9 | 24.01 |
14.9 | 14-14.9 | 34.81 |
14.9 | 15-14.9 | 0.01 |
14.9 | 10-14.9 | 24.01 |
14.9 | 9-14.9 | 34.81 |
14.9 | 20.14.9 | 26.01 |
14.9 | 21-14.9 | 37.21 |
14.9 | 19-14.9 | 16.81 |
14.9 | 15-19.5 | 0.01 |
14.9 | 16-14.9 | 1.21 |
14.9 | 17.14.9 | 4.41 |
14.9 | 13-14.9 | 3.61 |
14.9 | 13-14.9 | 3.61 |
14.9 | 12-14.9 | 8.41 |
14.9 | 10-14.9 | 24.01 |
14.9 | 9-14.9 | 34.81 |
14.9 | 20-14.9 | 26.01 |
14.9 | 14-14.9 | 0.81 |
14.9 | 16-14.9 | 2.21 |
14.9 | 17-14.9 | 4.41 |
14.9 | 18-.14.9 | 9.61 |
14.9 | 19-14.9 | 16.81 |
14.9 | 21-14.9 | 37.21 |
14.9 | 10-14.9 | 24.01 |
14.9 | 9-14.9 | 34.81 |
14.9 | 17-14.9 | 4.41 |
14.9 | 19-14.9 | 16.81 |
14.9 | 21-14.9 | 37.21 |
14.9 | 11-14.9 | 15.21 |
14.9 | 12-14.9 | 8.41 |
SUM=514.7 |
SD=SQ MEAN
SD =22.69
Question 4. Is this data from a normal distribution? Why?
This sample is not from a normal distribution. For data to be considered as a normal distribution, the mean, mode, and median must be equal. In the above case, the mean is 14.9, the mode is 10, and the median is 14.5, indicating that the data is not from a sample of normal distribution. Also, for data to be considered as a normal distribution, the histogram must mimic the S-shaped curve.
Question 5. Determine the Z score A Percentile of a Student Who Studied 22 Hours in a Week
Where X=22
Mean=14.9
S D= 22.69
Then Z = 22-14.9/22.69
Z score = 0.313
Percentile
From the normal distribution table, the Z value lies in the 75th percentile. Thus, this implies that 75% of the observation are distributed around the mean
PART B: CRITICAL THINKING
Question 6. Given the mean of 10 and a median of 14, illustrate if the statement follows or does not follow
Data is normally distributed-Does not follow
Mode of data is 10-Does not follow
Data is skewed to the left-Follow
The standard deviation of the data is 4- Does not follow.
Worksheet for Chapter 6
Part A: Equations and Proportions
Question 1.a)
2/y+5=-3/y-6
2(y-6)=-3(y+5)
2y-12=-3y-15
2y+3y=-15+12
Y=-0.6
Question 1.b)
Solve for S=a1/1-r
(1-r)S=a1
(1-r) = a1/S
1-r=a1/S
R=a1/S+1
Question 1 C)
7*(3X-2)+5 =6*(2X-1)+24
21X – 14+5=12X – 6+24
21X-12X=-6+24+14-5
9X=27
X=3
PART B: Word Problems
Question 1
Weight on earth=360 pounds
Weight on moon =60 pounds
If 360 E= 60 M
196 E =?
= 60*196/360
Weight on Moon is equal to 32.67 pounds
Question 2
A | B | ||
Annual membership | 300 | 40 | |
List Premium | 70% | 90% |
Let Manufacture price be X
For Plan A
For plan B
Number of Merchandize to purchase to pay same price under both plans, then Plan A must be equal to Plan B
300+0.7X=40+0.9X
300 – 40=0.9X-0.7X
260=0.2X
X=1300
Hence, one must purchase 1,300 units for price of both plans to be equal. At this number of units, for plan A you will pay
300+0.7*1300
=1,210
Similarly for Plan B
40+ 0.9*1300
1,210
Question 3
P=42-0.5X
IF P=5%
Then;
0.05=42-0.5X
0.5X=42-0.05
0.5X=41.95
X=83.9
Hence, after 83.9 years, less than 5% of Americans will be smoking cigarettes. The data is calculated from 1965 onwards.
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