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## Description

SESSION |
FEBRUARY/MARCH 2024 |

PROGRAM |
MASTEROF COMPUTER APPLICATIONS (MCA) |

SEMESTER |
III |

COURSE CODE & NAME |
DCA7101– PROBABILITY AND STATISTICS |

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** **

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**Set-I**

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**In a certain college, 4% of men students and 1% of women students are taller than 1.8m. Furthermore, 60% of the students are women. If a student is selected at random and is found taller than 1.8m, what is the probability that the student is a woman ?**

**Ans 1.**

Let’s start by identifying the given information:

- 4% of men students are taller than 1.8m.
- 1% of women students are taller than 1.8m.
- 60% of the students are women.

We can represent the total number of students as a circle, and divide it into two parts: men and women.

The total number of students is not given, but we can use the information that 60% of the students are women to find the number of women. Let’s say the total number of women is 60% of x, where x is the total number of students. Then, the number of women is 0.6x.

Similarly, the number of men is x –

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**Find the constant k so that**

**(****, ****) = (**** + 1)****, 0 <****< 1, ****> 0 **

**0, elsewhere**

**is a joint probability density function. Are X and Y independent**

**Ans 2.**

First, let’s break down the given function (, ) = ( + 1)e^x. We have a product of two terms: ( + 1) and e^x. The exponential function e^x is a probability density function of a continuous random variable X, which has a support on the interval (0, ∞). The constant k is being multiplied by (x + 1), which is a function of x.

To make this a joint probability

**Calculate the arithmetic mean by means of Step deviation method:**

Marks |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |

No. of Students |
20 |
24 |
40 |
36 |
20 |

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**Ans 3.**

To calculate the arithmetic mean using the Step Deviation method, we’ll follow these steps:

Step 1: Calculate the total number of students (N) and the total marks (T).

N = 20 + 24 + 40 + 36 + 20 = 140

T = ?

Step 2: Calculate the step size (h). The step size is the difference between the upper limit of a class interval and the lower limit of the previous class interval.

h = 10 (since the interval is from 0-10, 10-20, 20-30, etc.)

Step 3: Calculate the frequency of each class interval and multiply it by the mid-point of each interval.

For example, for the interval 0-10, the mid-point is (0 + 10) / 2 = 5. The frequency is 20, so we multiply 5 by 20 to get:

20 × 5 = 100

Repeat this process for each interval:

Interval Mid-point

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**Set-II**

**Find Karl Pearson’s correlation coefficient for the data is**

X |
20 |
16 |
12 |
8 |
4 |

y |
22 |
14 |
4 |
12 |
8 |

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**Ans 4.**

Karl Pearson’s correlation coefficient, also known as Pearson’s r, is a measure of the linear correlation between two continuous variables X and Y. It ranges from -1 (perfect negative correlation) to 1 (perfect positive correlation), with 0 indicating no correlation.

To calculate Pearson’s r, we need to follow these steps:

- Calculate the means of X and Y:

X̄ = (20 + 16 + 12 + 8 + 4

** **

**Let [X,Y] be an absolutely continuous random vector with support****= {(****,****): 0 ≤****≤****≤ 2} i.e. ,**

**is the set of all couples (x,y) such that 0 ≤ **** ≤ 2&0 ≤ **** ≤ ****. Let the joint -probability density function of [X,Y] be**

**0, otherwise **

** **

**Ans 5.**

**Problem Statement:**

We have a random vector [X, Y] with support R, which is the set of all couples (x, y) such that:

0 ≤ x ≤ y ≤ 2

The joint probability density function (PDF) of [X, Y] is given by:

f_xy(X, Y) = 3/8y, if (x, y

**In a random sample of 100 persons taken from village A, 60 are found to be consuming tea. In another sample of 200 persons taken from village B, 100 persons are found to be consuming tea. Do the data reveal significant difference between the two villages so far as the habit of taking tea is concerned?**

**Ans 6.**

To determine if there’s a significant difference between the two villages, we need to analyze the data using a statistical method. In this case, we can use a hypothesis test.

**Step 1: State the null and alternative hypotheses**

**Null hypothesis (H0):** There is no significant difference in the proportion of people consuming tea