Question

A chip used in the production of particleboard has a 15% chance of containing excessive bark. What is the probability of having more than 860 bark-free chips in a batch of 1,000?  

Answer 1

Given =

Probability [ particleboard contain excessive bark] = 0.15

probability [ barkfree chip in particle board] = 1 - 0.15 = 0.85 = p

The population proportion of success is p = 0.85,

and the sample size is n= 1000.

We need to compute Pr(X≥860):

Generally, we use binomial distribution in order to calculate probability in this scenario. but since the sample size is larger here we can make use of normal approximation of binomial distribution to calculate the required probability.

The population mean is computed as:

μ=n⋅p=1000⋅0.85=850

and the population standard deviation is computed as:

$\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{ 1000 \cdot 0.85 \cdot (1-0.85)} = 11.2916$

Therefore, we get that

$\Pr( X_{Binomial} \geq 860 ) \approx \Pr( X_{Normal} \geq 859.5 )$

$\Pr \left(Z \ge \frac{ X_{Normal} - 850}{ 11.2916}\right)$

$= \Pr\left(Z \ge \frac{ 860 - 850}{ 11.2916}\right)$

$= \Pr(Z \ge 0.8413) = 1 - 0.7999 = 0.2$

so then, we conclude that

$Pr(X\geq 860)\approx 0.2$  

which means there is 20% chance of having more than 860 bark-free chips in a batch of 1,000.

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