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

At a store, it is known that

1

out of every

9

customers uses a gift voucher in part-payment for purchases. A randomly selected sample of

72

customers is taken. Use a suitable approximation and continuity correction to find the probability that at most

6

of these customers use a gift voucher in part-payment for their purchases.

Accepted Answer

Given that at a store, it is known that $1$ out of every $9$ customers uses a gift voucher in part-payment for purchases.

Let random variable $X$ is number of uses a gift voucher in part-payment for purchases.

Taking a sample of $72$ customers.

Let success is customer uses a gift voucher in part-payment for purchases.

$\Rightarrow p = \frac{1}{9}$

Here, $n = 72$

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

Here $X ~ B (72, \frac{1}{9})$

Probability of failure is $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$

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

Here, $n p = 72 \times \frac{1}{9} = 8 > 5$ ,

$1$

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

Mean $μ = n p = 8$ ,

Variance $σ^{2} = n p q = 72 \times \frac{1}{9} \times \frac{8}{9} = 7.11$

Probability that at most $6$ of these customers uses a gift voucher in part-payment for purchases is $P (X \leq 6)$ .

Using the continuity correction, standardising the value of $x = 6.5$ .

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

Here, $μ = 8, σ^{2} = 7.11$

$\Rightarrow z = \frac{6.5 - 8}{\sqrt{7.11}} \Rightarrow z = \frac{- 1.5}{2.666} = - 0.563$

So, $P (X \leq 6) = P (Z \leq - 0.563)$

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

$\Rightarrow P (X \leq 6) = 1 - ϕ (0.563) \Rightarrow P (X \leq 6) = 1 - 0.7133 \Rightarrow P (X \leq 6) = 0.2867$

Important Questions on The Normal Distribution

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