Name: Random Variables in Probability Theory
Uploaded: 2019-07-19
Duration: 8 min 57 s
Description: What do we mean by random variables? Watch this video to learn about random variables alongside continuous random variables and discrete random variables wit...

Question

An opinion poll was taken before an election. The table shows the percentage of voters who said they would vote for parties $A, B$ and $C$ .

Party	$A$	$B$	$C$
Votes $(%)$	$36$	$41$	$23$

Find an approximation for the probability that, in a random sample of $120$ of these voters exactly $50$ said they would vote for party $B$ .

Accepted Answer

Given that table of opinion poll, which was taken before election,

Party $A$ $B$ $C$

Votes $(%)$ $36$ $41$ $23$

Taking a random sample of $120$ of these voters.

So, $n = 120$

Let random variable $X$ is the number of voters, who said they would vote for party $B$ .

Let success is the voter said they would vote for party $B$ .

So, probability of success $p = 41 % = 0.41$

We know that for the binomial distribution $X ~ B (n, p)$ , $n$ is a number of trials and $p$ is the probability of success.

Here $X ~ B (120, 0.41)$

Probability of failure is $q = 1 - p = 1 - 0.41 = 0.59$

A binomial distribution can be well-approximated by a normal distribution if $n p$ and $n q$ are both greater than $5$ .

Here, $n p = 120 \times 0.41 = 49.2 > 5$ ,

$n q = 120 \times 0.59 = 70.8 > 5$

Hence, given binomial distribution can be well-approximated by a normal distribution.

Mean $μ = n p = 49.2$ ,

Variance $σ^{2} = n p q = 120 \times 0.41 \times 0.59 = 29.028$

Probability that exactly $50$ voters said they would vote for party $B$ is $P (X = 50)$ .

$P (X = 50) = P (49.5 < X < 50.5) \Rightarrow P (X = 50) = P (X < 50.5) - P (X < 49.5)$

Using the continuity correction, standardising the value of $x = 50.5$ and $x = 49.5$ .

Standardising value $z = \frac{x - μ}{σ}$

Here, $μ = 49.2, σ^{2} = 29.028$

For $x = 50.5$ ,

$\Rightarrow z = \frac{50.5 - 49.2}{\sqrt{29.028}} \Rightarrow z = \frac{1.3}{5.388} = 0.241$

For $x = 49.5$ ,

$\Rightarrow z = \frac{49.5 - 49.2}{\sqrt{29.028}} \Rightarrow z = \frac{0.3}{5.388} = 0.056$

So, $P (X = 50) = P (Z < 0.241) - P (Z < 0.056)$

We know that, if $z = a, P (Z < a) = ϕ (a)$

$\Rightarrow P (X = 50) = ϕ (0.241) - ϕ (0.056) \Rightarrow P (X = 50) = 0.5952 - 0.5223 \Rightarrow P (X = 50) = 0.0729$

Important Questions on The Normal Distribution

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An opinion poll was taken before an election. The table shows the percentage of voters who said they would vote for parties A, B and C. Party A B C Votes % 36 41 23 Find an approximation for the probability that, in a random sample of 120 of these voters exactly 50 said they would vote for party B.

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !