# Solved (Free): Suppose that 15 % of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts

#### ByDr. Raju Chaudhari

Mar 28, 2021

Suppose that 15 % of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the approximate probability that X is between 25 and 35 (inclusive of both end points)?

#### Solution

Here $X$ denote the number of shafts that are non-confirming and can be reworked.

Let $p$ be the probability that a shaft is non-confirming.

Given that $p=0.15$ and $n =200$. Thus $X\sim B(200, 0.15)$.

Since $n*p = 30 \geq 10$ and $n*(1-p)= 170\geq 10$, we use Normal approximation to Binomial distribution.

Mean is
 \begin{aligned} E(X)=\mu&= n*p \\ &= 200 \times 0.15 \\ &= 30. \end{aligned}

and standard deviation is

 \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{200 \times 0.15 \times (1- 0.15)}\\ &=5.0498. \end{aligned}

The approximate probability that X is between 25 and 35 (inclusive of both end points) is

 \begin{aligned} P(25\leq X\leq 35) &= P(24.5\leq X\leq 35.5)\\ &=P\bigg(\frac{24.5- 30}{5.0498}\leq \frac{X-\mu}{\sigma}< \frac{35.5- 30}{5.0498}\bigg)\\ &= P(-1.09 < Z < 1.09)\\ & =P(Z < 1.09)-P(Z < -1.09)\\ & = 0.8621-0.1379\\ & = 0.7242 \end{aligned}