AIOU 1429 Business Mathematics Solved Assignment 2 Spring 2025
AIOU 1429 Assignment 2
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AIOU 1429 Solved Assignment 1 Spring 2025
AIOU 1429 Business Mathematics Solved Assignment 1 Spring 2025
AIOU 1429 Assignment 1
Q1.(a There are 16 green, 20 red and 24 yellow balls in a basket. If we pick a ball at random what is the probability that
i. The ball is green.
ii. The ball is not green and red.
(20 Marks)
Step 1: Total Number of Balls
There are:
- 16 green balls
- 20 red balls
- 24 yellow balls
The total number of balls is 16 + 20 + 24 = 60.
Step 2: Calculating the Probabilities
i. Probability that the ball is green
- The number of green balls is 16.
- Total number of balls is 60.
-
The probability of picking a green ball is:
P(green) = 16 / 60, which simplifies to 4/15.
ii. Probability that the ball is not green and red
- This phrase means the ball is neither green nor red.
- Therefore, the ball must be yellow.
-
The number of yellow balls is 24.
The probability of picking a yellow ball is:
P(yellow) = 24 / 60, which simplifies to 2/5.
Final Answers:
- The probability that the ball is green is 4/15.
- The probability that the ball is not green and red (i.e., the ball is yellow) is 2/5.
Q1.(b Differentiate between continuous and discrete random variables with the help of examples.
Continuous Random Variables
A continuous random variable can take any value within a given range. It represents measurements and can have decimals or fractions.
Examples:
- Height of students in a class (could be 160 cm, 160.2 cm, 160.35 cm, etc.)
- Time taken to finish a race (could be 12.45 seconds, 12.467 seconds, etc.)
- Temperature in a city (e.g., 24.5°C, 24.51°C)
Since these values can be infinitely precise, they require intervals rather than exact numbers.
Discrete Random Variables
A discrete random variable takes specific, countable values. It represents things that are counted rather than measured.
Examples:
- Number of students in a class (can be 30, 31, 32—not 30.5!)
- Number of cars in a parking lot (e.g., 15, 20, 25—not fractions)
- Number of goals scored in a soccer match (e.g., 3 goals, 5 goals—never 3.7 goals!)
Since discrete variables come in distinct numbers, they can be listed or counted directly.
Summary of Key Differences
| Feature | Continuous Random Variable | Discrete Random Variable |
|---|---|---|
| Values | Can take any value within an interval | Specific, countable numbers |
| Nature | Measured (e.g., height, weight, temperature) | Counted (e.g., number of students, goals, cars) |
| Examples | Time, speed, distance | Coins flipped, books on a shelf |
Q2. The number of fire alarms pulled each hour fluctuates in Islamabad. The probability table of different alarms per hour is shown(20 Marks)
| No. of Alarm Pulled | Probability |
|---|---|
| Less than 8 | 0.12 |
| 8 | 0.24 |
| 9 | 0.28 |
| 10 | 0.26 |
| More than 10 | 0.10 |
a) Probability that more than 8 alarms will be pulled
- The probability of pulling 9 alarms is 0.28.
- The probability of pulling 10 alarms is 0.26.
- The probability of pulling more than 10 alarms is 0.10.
- The total probability is:
0.28 + 0.26 + 0.10 = 0.64.
b) Probability that the number of alarms pulled is between 8 and 9 (inclusive)
- The probability of pulling 8 alarms is 0.24.
- The probability of pulling 9 alarms is 0.28.
- The total probability is:
0.24 + 0.28 = 0.52.
Final Answers:
- The probability that more than 8 alarms will be pulled is 0.64.
- The probability that the number of alarms pulled is between 8 and 9 (inclusive) is 0.52.
Q3. The data on ocean storms in the USA for the last fifty years is given below(20 Marks)
| No. of Storms | Frequency |
|---|---|
| 0 | 2 |
| 1 | 7 |
| 2 | 10 |
| 3 | 15 |
| 4 | 7 |
| 5 | 12 |
| 6 | 7 |
| Total: 60 | |
a) Construct probability distribution for this data.
The probability of each event occurring is given by:
P(X) = Frequency / Total Count
Since the total number of observations is 60, we compute probabilities for each storm count:
| No. of Storms (X) | Frequency (f) | Probability P(X) |
|---|---|---|
| 0 | 2 | 2/60 = 0.0333 |
| 1 | 7 | 7/60 = 0.1167 |
| 2 | 10 | 10/60 = 0.1667 |
| 3 | 15 | 15/60 = 0.2500 |
| 4 | 7 | 7/60 = 0.1167 |
| 5 | 12 | 12/60 = 0.2000 |
| 6 | 7 | 7/60 = 0.1167 |
b) Draw a histogram for this distribution.
Q4.(a Solve the second degree equation and find the nature of its roots.
y^2-y-2=0(10 Marks)
Q4.(b Solve the inequality and represent the solution on the real line.(10 Marks)
Q5.(a Find the point of intersection of the following lines if it exists.
x + 2y = 3
2x - y = 1(10 Marks)
Let's solve for the intersection point by solving the system of equations:
x + 2y = 3
2x - y = 1
Step 1: Express One Variable in Terms of the Other
From the first equation, solve for x:
x = 3 - 2y
Step 2: Substitute in the Second Equation
Replace x in the second equation:
2(3 - 2y) - y = 1
Expanding:
6 - 4y - y = 1
6 - 5y = 1
Step 3: Solve for y
-5y = -5
y = 1
Step 4: Solve for x
Substituting y = 1 into x = 3 - 2y:
x = 3 - 2(1) = 1
Conclusion:
The lines intersect at (1,1).
Q5.(b Let C mean Celsius degree and F mean Fahrenheit temperature scale. Find a linear equation for C if its slope is 5/9 and the C-intercept is -100/9.(10 Marks)
To find the linear equation for C in terms of F, we use the standard equation of a line:
C = mF + b
where:
m is the slope (5/9)
b is the C-intercept (-100/9)
Substituting these values:
C = (5/9)
F - (100/9)
Conclusion:
The required linear equation relating Celsius (C) to Fahrenheit (F) is:
C = (5/9)
F - (100/9)