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

17 %

of the people interviewed in a survey said they watch more than two hours of

TV

per day. A random sample of

300

of those who were interviewed is taken. Find an approximate value for the probability that at least one-fifth of those in the sample watch more than two hours of

TV

per day.

Accepted Answer

Given that $17 %$ of the people interviewed in a survey said they watch more than two hours of $300$ per day.

Taking a random sample of $300$ people who were interviewed.

So, $n = 300$

Let random variable $X$ is the number of people, who were watch more than two hours of $TV$ per day.

Probability that randomly selected people who was watch more than two hours of $TV$ per day $= 17 % = \frac{17}{100} = 0.17$

So, $p = 0.17$

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 $X ~ B (300, 0.17)$

Probability of failure is $q = 1 - p = 1 - 0.17 = 0.83$

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

Here, $n p = 300 \times 0.17 = 51 > 5$ ,

$n q = 300 \times 0.83 = 249 > 5$

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

Mean $μ = n p = 51$ ,

Variance $σ^{2} = n p q = 300 \times 0.17 \times 0.83 = 42.33$

Probability that at least one-fifth of those in the sample watch more than two hours of $TV$ per day is $P (X \geq \frac{300}{5})$ .

$P (X \geq \frac{300}{5}) = P (X \geq 60)$

Using the continuity correction, standardising the value of $x = 59.5$ .

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

Here, $μ = 51, σ^{2} = 42.33$

$\Rightarrow z = \frac{59.5 - 51}{\sqrt{42.33}} \Rightarrow z = \frac{8.5}{6.506} = 1.306$

So, $P (X \geq 60) = P (Z \geq 1.306)$

We know that, if $z = a, P (Z \geq a) = 1 - ϕ (a)$

$\Rightarrow P (X \geq 60) = 1 - ϕ (1.306) \Rightarrow P (X \geq 60) = 1 - 0.9042 \Rightarrow P (X \geq 60) = 0.0958$

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

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

$17 %$ of the people interviewed in a survey said they watch more than two hours of $TV$ per day. A random sample of $300$ of those who were interviewed is taken. Find an approximate value for the probability that at least one-fifth of those in the sample watch more than two hours of $TV$ per day.

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !

17% of the people interviewed in a survey said they watch more than two hours of TV per day. A random sample of 300 of those who were interviewed is taken. Find an approximate value for the probability that at least one-fifth of those in the sample watch more than two hours of TV per day.

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !

$17 %$ of the people interviewed in a survey said they watch more than two hours of $TV$ per day. A random sample of $300$ of those who were interviewed is taken. Find an approximate value for the probability that at least one-fifth of those in the sample watch more than two hours of $TV$ per day.