Question

(a)

The records show that 8% of the items produced by a machine do not meet the specifications. You take a sample of 100 units. Find the standard deviation (Use exactly two decimal places)

(b)

The records show that 8% of the items produced by a machine do not meet the specifications. You take a sample of 100 units. What is the probability that this sample of 100 units contains five or more defective units?Use the normal approximation to the binomial distribution to answer the question

(c)

The records show that 8% of the items produced by a machine do not meet the specifications. Use the normal approximation to the binomial distribution to answer the following questions. What is the probability that a sample of 100 units contains. What is the probability that this sample of 100 units contains ten or fewer defective units? Use the normal approximation to the binomial distribution to answer the question.

Answer 1

a) The standard deviation here is computed as:

$\LARGE \sigma = \sqrt{np(1-p)}$

$\LARGE \sigma = \sqrt{100*0.08*0.92} = 2.7129$

b) The number of defective parts out of 100 could be modelled here as:

$\LARGE X \sim Bin(n= 100, p = 0.08)$

This can be approximated to a normal distribution as:

$\LARGE X \sim N(\mu = np, \sigma = 2.7129)$

$\LARGE X \sim N(\mu = 8, \sigma = 2.7129)$

The required probability here is computed as:

$\LARGE P(X \geq 5)$

Applying the continuity correction, we get here:

$\LARGE P(X > 4.5)$

Converting this to a standard normal variable, we get:

$\LARGE P(Z > \frac{4.5 - 8}{2.7129})$

$\LARGE P(Z > -1.29)$

Getting it from the standard normal tables, we get:

$\LARGE P(Z > -1.29) = 0.9015$

Therefore 0.9015 is the required probability here.

c) The required probability here is computed as:

P( X <= 10 )

Applying the continuity correction, we get here:

P(X < 10.5 )

Converting this to a standard normal variable, we get:

$\LARGE P(Z < \frac{10.5 -8}{\sqrt{7.36}})$

$\LARGE P(Z < 0.92)$

Getting it from the standard normal tables, we get:

$\LARGE P(Z < 0.92) = 0.8216$

Therefore 0.8216 is the required probability here.