Question

The mean and standard deviation for the diameter of a certain type of steel rod are mu = 0.503 cm and sigma = 0.03cm. Let X denote the average of the diameters of a batch of 100 such steel rods. The batch passes inspection if Xbar falls between 0.495 and 0.505cm.

1. What is the approximate distribution of Xbar? Specify the mean and the variance and cite the appropriate theorem to justify your answer.

2. What is the approximate probability the batch will pass inspection?

3. Over the next six months 40 batches of 100 will be delivered. Let Y denote the number of batches that will pass inspection.

(a) the distribution of Y is: Binomial, hypergeometric, negative binomial, OR poisson?

(b) give the approximation, as accurately as possible, to the probability P(Y ≤ 30).

Answer 1

1)for n=100 is greater then 30 ; therefore from central limit theorum;

approximate distribution of Xbar is normal with

mean =0.503

and variance=²/n=0.03²/100=0.000009

2)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	0.503
std deviation   =σ=	0.0300
sample size       =n=	100
std error=σx̅=σ/√n=	0.0030

approximate probability the batch will pass inspection:

probability =

P(0.495<X<0.505)

=

P(-2.67<Z<0.67)=

0.7486-0.0038=

0.7448

3)

a)   the distribution of Y is: Binomial with paramter n=40 and p=0.7448

b)

n=	40	p=	0.7448
here mean of distribution=μ=np=		29.792
and standard deviation σ=sqrt(np(1-p))=		2.7573
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

probability P(Y ≤ 30).:

probability =

P(X<30.5)

=

P(Z<0.26)=

0.6026