Class 11 Notes for all pakistani boards Statistics Chapter 7 (RandomVariable and Probability Distribution).
Statistics Chapter 7 (RandomVariable and Probability Distribution)
Table of Contents
Q.1 Define random variables and its probability distribution. Explain by means of two examples.
Answer:
RANDOM VARIABLES:
If you have ever taken an algebra class, you probably learned about different variables like x, y and maybe event. Some examples of variables include x = number of heads or y = number of cell phones or z = running time of movies. Thus, in basic math, a variable is an alphabetical character that represents an unknown number.
Well, in probability, we also have variables, but we refer to them as random variables. A random variable is a variable that is subject to randomness, which means it can take on different values.
As in basic math, variables represent something, and we can denote them with an x, a y, or any other letter for that matter. However, in statistics, it is normal to use an X to denote a random variable. The random variable takes on different values depending on the situation. Each value of the random variable has a probability or percentage associated with it.
Example
Two dice are rolled and we define the familiar sample space
Ω = {(1, 1), (1, 2) … (6, 6)} containing 36 elements.
Let X denote the random variable whose value for any element of Ω is the sum of the numbers on the two dice. Then the range of X is the set containing the 11 values of X: 2,3,4,5,6,7,8,9,10,11,12.
Each ordered pair of Ω has associated with it exactly one element of the range as required by Definition 1. However, in general, the same value of X arises from many different outcomes.
For example X (ok) = 5 is any one of the four elements of the event
{(1, 4), (2, 3),(3, 2),(4, 1)}
PROBABILITY DISTRIBUTION OF RANDOM VARIABLE:
Q.2 Indicate whether the following examples are discrete or continuous random variables. If the variable discrete state whether it takes fractional values or only integers values.
i) The G.N.P of Pakistan
ii) The shoe size of randomly selected European man.
iii) The time taken by a flight from Peshawar to Karachi in hours.
iv) The number of flight between Peshawar to Karachi in a 24-hours period.
v) The price of gallon of gasoline in Rs. 6.
vi) The number of automobile accidents per year in Peshawar.
vii) The length of time to play 18 holes of golf.
viii) The amount of milk produced yearly by a particular cow.
ix) The number of eggs lay each month by a hen.
x) The number of building permits issued each month in a certain city.
xi) The weight of grain produced per acre.
Read more: Statistics Chapter 6 (Probability) | Class 11 for KPK
Answer:
i) The G.N.P of Pakistan (DISCRETE VARIABLE)
ii) The shoe size of randomly selected European man. (DISCRETE VARIABLE)
iii) The time taken by a flight from Peshawar to Karachi in hours.(CONTINUOUS VARIABLES)
iv) The number of flight between Peshawar to Karachi in a 24-hours period. (DISCRETE VARIABLE)
v) The price of gallon of gasoline in Rs. 6. (DISCRETE VARIABLE)
vi) The number of automobile accidents per year in Peshawar. (DISCRETE VARIABLE)
vii) The length of time to play 18 holes of golf.(CONTINUOUS VARIABLES)
viii) The amount of milk produced yearly by a particular cow.(CONTINUOUS VARIABLES)
ix) The number of eggs lay each month by a hen. (DISCRETE VARIABLE)
x) The number of building permits issued each month in a certain city. (DISCRETE VARIABLE)
xi) The weight of grain produced per acre.(CONTINUOUS VARIABLES)
Q.9) A random variable x has possible values x = 1, 2, 3, 4. Indicate whether or not each of following is a probability function for such a variable is met, only met.
a) P(1) =0.05, P(2) =0.35, P(3) =0.40, P(4) =0.20
b) P(1) =1/3, P(2) =½, P(3) = ¼, P(4) =1/16
c) P(1) =12%, P(2) =60%, P(3) =18%, P(4) =10%
d) P(1) =0.66, P(2) =0.30, P(3) = -0.2, P(4) =0.1
e) P(1) =2, P(2) =3, P(3) =4, P(4) =1
Answer:
a) P(1) =0.05, P(2) =0.35, P(3) =0.40, P(4) =0.20
P(1)+ P(2)+P(3)+P(4) =0.05+0.35+0.40+0.20=1
so Yes it met because sum of probability =1
b) P(1) =1/3, P(2) =½, P(3) = ¼, P(4) =1/16
P(1)+ P(2)+P(3)+P(4) =1/3+1/2+1/4+1/16=1.14
SO No it not met because sum of probability is not equal to 1
c) P(1) =12%, P(2) =60%, P(3) =18%, P(4) =10%
P(1)+ P(2)+P(3)+P(4) =0.12+0.60+0.18+.10=1
so Yes it met because sum of probability =1|
d) P(1) =0.66, P(2) =0.30, P(3) = -0.2, P(4) =0.1
No it not met because of negitive probability.
e) P(1) =2, P(2) =3, P(3) =4, P(4) =1
No the probability must be 0 and 1
Q.10) Explain the concept of random variable. Give some examples from daily life.
Answer:
Concepts of random variables and there construction from various fields:
When a coin is tossed, we may not be interested in the actual outcome head (H) or tail (T), but the values 0, 1 indicating the number of heads. Similarly, we may not be interested in “defective” or “not defective” part produced by an industrial process, but in values 0, 1 indicating “not defective” and “defective” parts respectively. These numerical outcomes can be summarized by a variable, called random variable. The values 0 and 1 are the variable values determined by the outcomes of the random experiments. Any variable quantity whose numerical values are determined by the’ outcomes of a random experiment is called a random variable.
Capital letters such as X, Y, Z are used to denote random variables and the corresponding small letters, such as x, y, z are used to denote values assumed by the random variables.
Their constructions depend upon the random experiments performed in various fields of inquiry- For example the number of crimes committed in a large city during one-year duration as having outcomes 0, 1, 2…. The number of defective component produced in a pair by an industrial
Process, which may assume any one of three values 0. 1. and 2.
For example, in a random variable giving the number of heads obtained. List the element of sample space S for three tosses of the coin and to each sample point assigns a value w of W.
Solution:
The sample space S consists of 23=8, sample point as given below
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
Let us consider the variable w, which is the number of heads. Then
Q.11 a) Define the following types of random variables
i) Discrete random variable.
ii) Continuous random variable.
b) Obtain the probability distribution of number of heads in two tosses of a coin.
Answer:
Discrete random variable:
A variable is a quantity whose value changes.
A discrete variable is a variable whose value is obtained by counting.
Examples:
· number of students present
· Number of red marbles in a jar
· Number of heads when flipping three coin
· Students’ grade level
A discrete random variable X has a countable number of possible values.
Example: Let X represent the sum of two dice.
Then the probability distribution of X is as follows:
| X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| P(X) | 1/ 36 | 2/ 36 | 3/ 36 | 4/ 36 | 5/ 36 | 6/ 36 | 5/ 36 | 4/ 36 | 3/ 36 | 2/ 36 | 1/36 |
To graph the probability distribution of a discrete random variable, construct a probability histogram.
PART B:
CONTINUOUS RANDOM VARIABLE:
A continuous random variable X takes all values in a given interval of numbers.
· The probability distribution of a continuous random variable is shown by a density curve.
· The probability that X is between an intervals of numbers is the area under the density curve between the interval endpoints
· The probability that a continuous random variable X is exactly equal to a number is zero
Q.13) Define a probability distribution, probability density function and distribution.
Answer:
Probability
Probability is the likelihood that an event will occur and is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails. Therefore, the probability of getting heads is 1 out of 2, or ½ (50%). Table 1.1 below shows the distribution of these probabilities:
Probability Distribution
Probability Distribution maps out the likelihood of multiple outcomes in a table or an equation. If we go back to the coin flip example, we already know that one flip of the coin has only two possible outcomes. However, if we flip the coin twice in a row, there are four possible outcomes (heads-heads, heads-tails, tails-heads, and tails-tails). So now that we have a series of potential outcomes, consider the probabilities of getting heads once, twice or zero times, as shown in Table 1.2 below:
The table above shows the distribution of probabilities for each possible result of getting heads in two flips of the coin, thus the table is called probability distribution. Notice that the two middle rows both reflect the probability of getting just one heads in both flips, so these two probabilities can be combined and rewritten as a probability of 2 out of 4 (50%), as in Table 1.3 below:
Probability Density Function
Most often, the equation used to describe a continuous probability distribution is called a probability density function. Sometimes, it is referred to as a density function, a PDF, or a pdf. For a continuous probability distribution, the density function has the following properties:
Since the continuous random variable is defined over a continuous range of values (called the domain of the variable), the graph of the density function will also be continuous over that range.
The area bounded by the curve of the density function and the x-axis is equal to one, when computed over the domain of the variable.
The probability that a random variable assumes a value between a and b is equal to the area under the density function bounded by a and b.
For example, consider the probability density function shown in the graph below. Suppose we wanted to know the probability that the random variable X was less than or equal to a. The probability that Xis less than or equal to a is equal to the area under the curve bounded by a and minus infinity – as indicated by the shaded area.
Note: The shaded area in the graph represents the probability that the random variable X is less than or equal to a. This is a cumulative probability. However, the probability that X is exactly equal to would be zero. A continuous random variable can take on an infinite number of values. The probability that it will equal a specific value (such as a) is always zero.
