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
