Question

2. Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be re- worked (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 a) At most 302 (2) b) Between 15 and 25 (both inclusive)? [2] c) Assume that the probability of at most I shafts being nonconforming is 0.9906. Find z. [3]

Answer 1

The number of steel shafts that are non confirming and can be reworked out of 200 shafts selected is modelled here as:

$X \sim Bin(n = 200, p =0.1)$

This is approximated to a normal distribution as:

$X \sim N(\mu = np, \sigma^2 = np(1-p ) )$

$X \sim N(\mu = 200*0.1, \sigma^2 = 200*0.1*0.9)$

$X \sim N(\mu = 20, \sigma^2 = 18 )$

a) The probability here is computed as:

P(X <= 30)

Applying the continuity correction, we have here:
P(X < 30.5)

Converting it to a standard normal variable, we have here:

$P(Z < \frac{30.5 - 20}{\sqrt{18}})$

Getting it from the standard normal tables, we have here:

Therefore 0.9933 is the required probability here.

b) The probability here is computed as:

P(15 <= X <= 25)

Applying the continuity correction, we have both sides here:

P( 14.5 < X < 25.5)

Converting it to a standard normal variable, we have here:

$P(\frac{14.5 -20}{\sqrt{18}} < Z < \frac{25.5 - 20}{\sqrt{18}})$

Getting it from the standard normal tables, we have here:

Therefore 0.8064 is the required probability here.

c) The probability of at most x shafts being non confirming is given to be 0.9906

Therefore, P(X <= x) = 0.9906

From standard normal tables, we have here:

P(Z < 2.349) = 0.9906

Therefore, we have here:

$\frac{x - 20}{\sqrt{18}} = 2.349$

$x - 20 = 2.349\sqrt{18}$

Therefore the value of x here is 29.