Question

Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (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. (Round your answers to four decimal places.) (a) What is the (approximate) probability that X is at most 30

Answer 1

Answer:

0.9726 = 97.26% approximate probability that X is at most 30

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).

This means that $p = 0.11$

Random sample of 200 shafts

This means that $n = 200$

Mean and Standard deviation:

$\mu = E(x) = np = 200*0.11 = 22$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.11*0.89} = 4.42$

(a) What is the (approximate) probability that X is at most 30

Using continuity correction, this is $P(X \leq 30 + 0.5) = P(X \leq 30.5)$, which is the pvalue of Z when X = 30.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 22}{4.42}$

$Z = 1.92$

$Z = 1.92$ has a pvalue of 0.9726.

0.9726 = 97.26% approximate probability that X is at most 30