NMIMS Decision Science Assignment solution for April 2025

1)  Calculate Quartile Deviation from the following

Earnings (Rs)Number of Persons
2525
2670
27210
28275
29430
30550
31340
32130
3390
3455
3525

SOLUTION:

Quartile Deviation (QD) is a measure of the spread or dispersion of a data set, specifically focusing on the middle 50% of the data. It is calculated using the difference between the third quartile (Q3) and the first quartile (Q1), then divided by two. The formula for Quartile Deviation is: QD= (Q3−Q1)/ 2

Where:

  • Q1 (First Quartile) represents the value below which 25% of the data points fall.
  • Q3 (Third Quartile) represents the value below which 75% of the data points fall.

The Quartile Deviation provides a simple way to understand the variability in a data set, especially when data contains outliers or extreme values. It is particularly useful when the data is not symmetrically distributed, as it excludes the extremes (outliers), focusing instead on the middle 50% of the data, making it less sensitive to outliers compared to other measures like the range or standard deviation.

In practice, Quartile Deviation is often used in fields such as economics, finance, and social sciences to analyze distributions and measure consistency or variation within a dataset. The smaller the Quartile Deviation, the more concentrated the data is around the median.

Total Number of Persons = 2200


Median (Q2) Calculation:

  • Q2 (Median) = N/ 2=2200/2 = 1100
  • Q2 lies between earnings Rs 29 and Rs 30.

Q1 Calculation:

  • Q1 = N4=2200/ 4=550
  • Q1 lies between earnings Rs 28 and Rs 29.

Q3 Calculation:

  • Q3 = 3N4=3×2200/4=1650
  • Q3 lies between earnings Rs 30 and Rs 31.

Quartile Deviation (QD):

QD= (Q3−Q1)/ 2=(30.5−28.5)/ 2=2/ 2=1

Thus, the Quartile Deviation (QD) is 1.

2)  Calculate Median from the following

Class – IntervalNumber of Students
0-105
10-208
20-307
30-4012
40-5028
50-6020
60-7010
70-8010

SOLUTION:

Step 1: Organize the Data

Class IntervalFrequency (f)Cumulative Frequency (CF)
0 – 1055
10 – 20813
20 – 30720
30 – 401232
40 – 502860
50 – 602080
60 – 701090
70 – 8010100

Step 2: Find the Median Class

The total number of students (N) is 100. The median corresponds to the value at the N/2 = 100/2 = 50

The cumulative frequency just before 50 is 32 (at the class interval 30-40), and the next cumulative frequency is 60 (at the class interval 40-50). Therefore, the Median Class is 40-50.

Step 3: Apply the Median Formula

The formula to calculate the median for grouped data is:

Where:

  • L = Lower boundary of the median class = 40
  • N = Total number of students = 100
  • CF = Cumulative frequency before the median class = 32
  • f= Frequency of the median class = 28
  • h = Class width = 10

Step 4: Calculate the Median

Substitute the values into the formula: Median=40+(50−32/ 28)×10= 40+(2850−32​)×10

Median = 40 + (18/28) *100

Median = 40+ 6.43 = 46.43

Final Answer:

The Median is 46.43.

3.a) A bag contains 6 white and 4 black balls. Two balls are drawn at random one after another without replacement. Find the probability that both drawn balls are white. (5 Marks)

SOLUTION:

“Probability – Conditional Probability without Replacement.”

Explanation:

  1. Probability:
    • Probability is a branch of mathematics that deals with calculating the likelihood or chance of an event occurring.
    • It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
  2. Conditional Probability:
    • Conditional probability is the probability of an event occurring given that another event has already occurred.
    • In this question, the events are dependent because the balls are drawn without replacement, meaning the probability changes after each draw.
  3. Without Replacement:
    • “Without replacement” means that once an item (in this case, a ball) is selected, it is not returned to the set of items available for future selections.
    • This makes the events dependent, as the outcome of the first draw affects the probability of the second draw.

In the given problem:

  • There are two events: drawing the first white ball and drawing the second white ball.
  • The probability of each event is calculated based on the remaining number of balls, which changes after each draw (since we are drawing without replacement).

This concept falls under the broader category of “Dependent Events” in probability, where the outcome of one event affects the probability of the next event.

To solve this, we need to find the probability that both drawn balls are white when two balls are drawn without replacement.

Step 1: Total Number of Balls

The total number of balls in the bag is:6 (white balls)+4 (black balls)=10 balls

Step 2: Probability of Drawing the First White Ball

The probability of drawing a white ball on the first draw is the ratio of white balls to the total number of balls: P(First white ball)=6/10=3/5

Step 3: Probability of Drawing the Second White Ball

After drawing the first white ball, there are now 5 white balls left and 9 balls remaining in total. So, the probability of drawing a white ball on the second draw is: P(Second white ball)=5/9

Step 4: Calculate the Joint Probability

Since the events are dependent (because the balls are drawn without replacement), we multiply the probabilities of each event: P(Both balls are white)=P(First white ball)×P(Second white ball) P(Both balls are white)=3/ 5×5/ 9=15/ 45=1/3

Final Answer:

The probability that both drawn balls are white is 1/3

3.b) In an intelligence test administered on 1000 children, the average was 42 and standard deviation was 24. Find the number of children exceeding a score of 50.   (5 Marks)

SOLUTION:

We use the z-score in this case because we are dealing with a normal distribution (as implied by the problem statement) and we want to find the probability of a specific score (50) in a distribution characterized by the mean and standard deviation. The z-score allows us to standardize the score and compare it to the normal distribution.

Why Z-Score is Used:

  1. Standardization:
    • The z-score helps us convert a specific score into a standardized value that tells us how many standard deviations away from the mean the score is.
    • This is important because the normal distribution is symmetric and characterized by its mean and standard deviation, but we often need to compare different scores within that distribution to understand their relative standing.
  2. Comparison to the Normal Distribution:
    • Once we have the z-score, it becomes easier to look up the cumulative probability in a z-table or use a standard normal distribution to determine the proportion of values below or above the given score.
    • In this case, we want to know the probability of scoring greater than 50, which is equivalent to the area under the normal curve to the right of the z-score corresponding to 50.
  3. Simplifies Probability Calculation:
    • Using the z-score makes it easier to compute probabilities in any normal distribution, regardless of the specific mean or standard deviation.
    • Without the z-score, we would need to directly work with the original distribution, which is more complex and inefficient for comparing different values.
Z Score formulae

To solve this, we need to find the number of children exceeding a score of 50 in an intelligence test. Given the following information:

  • Average score (Mean, μ) = 42
  • Standard deviation (σ) = 24
  • Total number of children = 1000
  • Score threshold = 50

We will use the z-score formula to standardize the score and find the probability of a child exceeding a score of 50.

Step 1: Calculate the Z-Score for 50

The z-score is calculated using the formula:

Step 2: Find the Probability for Z = 0.33

Using a z-table or a standard normal distribution table, we find the cumulative probability corresponding to a z-score of 0.33. From the table, the cumulative probability for a z-score of 0.33 is approximately:

This represents the probability of a child scoring less than 50.

Step 3: Find the Probability of Exceeding a Score of 50

The probability of a child exceeding a score of 50 is the complement of the probability of scoring less than 50:

Step 4: Calculate the Number of Children Exceeding a Score of 50

Since the number of children must be an integer, we round this to 371.

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