Question

A company that produces fine crystal knows from experience that 10% of its goblets have cosmetic flaws and must be classified as ”seconds”. If 20goblets are randomly selected, use Rand Table to get (a) what is the probability that exactly 2 are seconds? (b) what is the probability that at least 2 are seconds? (c) Find the mean number and the standard deviation of goblets that are classified as ”secondsd) Now suppose a random sample of 400 gobletsis selected. Use a normal approximation and R to find the probability that at least 50 are seconds

Answer 1

This is a binomial distribution question with
n = 20
p = 0.1
q = 1 - p = 0.9
where
$\small P(X = x) = \binom{n}{x} p^x q^{n-x}$

a) x = 2
$\small P(X = 2) = \binom{20}{2} 0.1^{2} 0.9^{20-2}$
P(X = 2) = 0.2852
b) x = 2
$\small \\P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \\P(X \ge 2) = 0.6084$
c) Since we know that
$\small \\Mean (\mu) = np = 2.0 \\Variance (\sigma^2) = npq = 1.8 \\Standard\; deviation (\sigma) = \sqrt{npq} = 1.3416$

This is a binomial distribution question with
n = 400
p = 0.1
q = 1 - p = 0.9
This binomial distribution can be approximated as Normal distribution since
np > 5 and nq > 5
Since we know that
$\small \\Mean (\mu) = np = 40.0 \\Standard\; deviation (\sigma) = \sqrt{npq} = 6.0$
Also
$\small z_{ score } = \frac{x-\mu}{\sigma}$
d) x = 50
P(x > 50.0)=?
$\small z = \frac {50.0-40.0}{6.0}$
z = 1.6667
This implies that
P(x > 50.0) = P(z > 1.6667) = 0.0478
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