# Random Samples

Contents

## Summary

• Random Samples is known as the collection of independent and identically distributed random variables such as ${ X }_{ 1 },\quad { X }_{ 2 },\quad { X }_{ 3 },\quad { X }_{ 4 }\quad .......\quad { X }_{ n }$
• Sample mean is considered a normal distribution which is represented as: $\bar { x } \quad \sim \quad N(\mu ,\quad \frac { { \sigma }^{ 2 } }{ n } )$
• The expected value also known as mean can be calculated as:
$E(\bar { X } )\quad =\quad E(\frac { S{ X }_{ i } }{ n } )\quad =\quad \frac { E({ X }_{ 1 })\quad +\quad E({ X }_{ 2 })\quad +\quad E({ X }_{ 3 })\quad +\quad ........\quad E({ X }_{ n }) }{ n }$
• Variance is a measurement of the spread between numbers in a data set. It can be calculated as: ${ s }^{ 2 }\quad =\quad \frac { 1 }{ n } \quad =\quad \sum _{ i\quad =\quad 1 }^{ n }{ { ({ x }_{ 1 }\quad -\quad \bar { x } ) }^{ 2 } }$

Random Samples is known as the collection of independent and identically distributed random variables such as ${ X }_{ 1 },\quad { X }_{ 2 },\quad { X }_{ 3 },\quad { X }_{ 4 }\quad .......\quad { X }_{ n }$.

#### Sample mean

We are already aware of the fact that mean is a measure of central tendency, and we also know that we inorder to calculate mean e.g if asked to calculate the mean height of students in 5th grade at a school. We will take height of all the students in the 5th grade at the school, add them all up and divide by the total number to get the mean. However, this way of finding mean can be nearly impossible if there are hundreds of students, so what we do is that we take a group of students from the grade (that is known as taking a sample) and calculate their height and take it’s mean. This is known as sample mean and It is considered to be more practical to measure a smaller sample from the set.

We use the following formula:

$\bar { x } \quad =\quad \frac { { X }_{ 1 }\quad +\quad { X }_{ 2 }\quad +\quad { X }_{ 3 }\quad +\quad { X }_{ 4 }\quad ........\quad { X }_{ n } }{ n }$

Moreover, if we draw sample of size n from a normal population then the distribution of sample mean is considered a normal distribution which is represented as:

$\bar { x } \quad \sim \quad N(\mu ,\quad \frac { { \sigma }^{ 2 } }{ n } )$

#### Example#1

Q. The masses of students at a college is a normal distribution $\bar { x } \quad \sim \quad N(70,\quad { 5 }^{ 2 })$. Four students are chosen at random, find the probability that their mean is less than 65kg.

Solution:

Step 1

We know that n = 4 so we can write our normal distribution as:

$\bar { x } \quad \sim \quad N(70,\quad \frac { { 5 }^{ 2 } }{ 4 } )$

Step 2

We can now say:

$P(\bar { X } \quad <\quad \frac { 65\quad -\quad 70 }{ \sqrt { \frac { 25 }{ 4 } } } )\quad =\quad P(Z\quad <\quad \frac { -1 }{ \frac { 5 }{ \sqrt { 4 } } } )$

$=\quad 1-\quad 0.9772\quad =\quad 0.00228$

#### Example #2

Q. The distribution of a random variable x is $\bar { x } \quad \sim \quad N(25,\quad 340)$.

Sample size n is drawn find the value of n if $P(\bar { x } \quad >\quad 28)\quad =\quad 0.005$.

Solution:

$P(\bar { x } \quad >\quad 28)\quad =\quad 0.005$

$P(\bar { x } \quad <\quad 28)\quad =\quad 1\quad -\quad 0.005$

$=\quad 0.995$

Looking at the value from the normal distribution table we can tell that 0.995 = 2.40

We can thus write:

$Z\quad =\quad \frac { \bar { x } \quad -\quad \mu }{ \frac { \sigma }{ \sqrt { n } } }$

$2.4\quad =\quad \frac { 28\quad -\quad 25 }{ \sqrt { \frac { 340 }{ n } } }$

By solving this equation we get:

n = 217.6      Ans

#### Expectation and Variance

We often refer to the expected value as the mean, and denote E(X) by µ for short. The mean can be calculated by using the formula written below:

$E(\bar { X } )\quad =\quad E(\frac { S{ X }_{ i } }{ n } )\quad =\quad \frac { E({ X }_{ 1 })\quad +\quad E({ X }_{ 2 })\quad +\quad E({ X }_{ 3 })\quad +\quad ........\quad E({ X }_{ n }) }{ n }$

Variance is a measurement of the spread between numbers in a data set. The variance measures how far each number in the set is from the mean. In simpler words it’s just the difference between an expected and actual result

Variance can be calculated by using the following formula:

${ s }^{ 2 }\quad =\quad \frac { 1 }{ n } \quad =\quad \sum _{ i\quad =\quad 1 }^{ n }{ { ({ x }_{ 1 }\quad -\quad \bar { x } ) }^{ 2 } }$