A chip used in the production of particleboard (Question 4) 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:

0.2

Step-by-step explanation:

Given that:

P(particleboard with excessive bark) = 0.15

The P(that particleboard do not have excessive bark) = 1 - 0.15

i.e the population proportion number of success p = 0.85

The sample size (n) is given to be = 1000

To find the P(X ≥ 860)

Since the sample size is large, we will apply the normal approximation of binomial distribution to treat this question.

The population mean [tex]\mu = n \times p[/tex]

[tex]\mu = 1000 \times 0.85[/tex]

[tex]\mu =850[/tex]

The population standard deviation [tex]\sigma = \sqrt{n*p(1-p)[/tex]

[tex]\sigma = \sqrt{1000*0.85*(1-0.85)}[/tex]

[tex]\sigma = \sqrt{1000*0.85*(0.15)}[/tex]

[tex]\sigma = \sqrt{127.5}[/tex]

[tex]\sigma = 11.2916[/tex]

Let X be the random variable which obeys a normal distribution;

Then;

[tex]P(X \ge 860) = P(Z\ge \dfrac{x- \mu}{\sigma})[/tex]

[tex]P(X \ge 860) = P(Z\ge \dfrac{860- 850}{11.2916})[/tex]

[tex]P(X \ge 860) = P(Z\ge \dfrac{10}{11.2916})[/tex]

P(X ≥ 860) = P(Z ≥ 0.8856)

P(X ≥ 860) = 1 - P(Z ≤ 0.8856)

From  z table

P(X ≥ 860) = 1 - 0.8122

P(X ≥ 860) = 0.1878

P(X ≥ 860) [tex]\simeq[/tex] 0.2

Thus, the probability of having more than 860 bark-free chips in a batch of 1,000 = 0.2

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