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

It is known that

2 %

of the cheapest memory sticks on the market are defective. In a random sample of

400

of these memory sticks, find approximately the probability that at least five but at most

11

are defective.

Accepted Answer

Given that $2 %$ of the cheapest memory sticks on the market are defective.

Taking a random sample of $400$ of these memory sticks.

So, $n = 400$

Let random variable $X$ is the number of memory sticks, which are defective.

Probability that randomly selected memory stick which is defective $= 2 % = \frac{2}{100} = 0.02$

Let success is the getting defective memory stick.

So, $p = 0.02$

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 $X ~ B (400, 0.02)$

Probability of failure is $q = 1 - p = 1 - 0.02 = 0.98$

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

Here, $n p = 400 \times 0.02 = 8 > 5$ ,

$n q = 400 \times 0.98 = 392 > 5$

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

Mean $μ = n p = 8$ ,

Variance $σ^{2} = n p q = 400 \times 0.02 \times 0.98 = 7.84$

Probability that at least five but at most $11$ memory sticks are defective is $P (5 \leq X \leq 11)$ .

$P (5 \leq X \leq 11) = P (4.5 \leq X \leq 11.5) \Rightarrow P (5 \leq X \leq 11) = P (X < 11.5) - P (X < 4.5)$

Using the continuity correction, standardising the value of $x = 11.5$ and $x = 4.5$ .

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

Here, $μ = 8, σ^{2} = 7.84$

For $x = 11.5$ ,

$\Rightarrow z = \frac{11.5 - 8}{\sqrt{7.84}} \Rightarrow z = \frac{3.5}{2.8} = 1.25$

For $x = 4.5$ ,

$\Rightarrow z = \frac{4.5 - 8}{\sqrt{7.84}} \Rightarrow z = \frac{- 3.5}{2.8} = - 1.25$

So, $P (5 \leq X \leq 11) = P (Z < 1.25) - P (Z < - 1.25)$

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

If $z = - a, P (Z < - a) = 1 - ϕ (a)$

$\Rightarrow P (5 \leq X \leq 11) = ϕ (1.25) - [1 - ϕ (1.25)] \Rightarrow P (5 \leq X \leq 11) = 0.8944 - 1 + 0.8944 \Rightarrow P (5 \leq X \leq 11) = 0.7888 \Rightarrow P (5 \leq X \leq 11) = 0.789$

It is known that $2 %$ of the cheapest memory sticks on the market are defective. In a random sample of $400$ of these memory sticks, find approximately the probability that at least five but at most $11$ are defective.

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !

It is known that 2% of the cheapest memory sticks on the market are defective. In a random sample of 400 of these memory sticks, find approximately the probability that at least five but at most 11 are defective.

Important Questions on The Normal Distribution

Learn and Prepare for any exam you want !

It is known that $2 %$ of the cheapest memory sticks on the market are defective. In a random sample of $400$ of these memory sticks, find approximately the probability that at least five but at most $11$ are defective.