Question

The breaking strength X of a certain rivet used in a machine engine has a mean 5000 psi and standard deviation 400 psi. A random sample of 25 rivets is taken. Consider the distribution of ?̅, the sample mean breaking strength.

(a) What is the probability that the sample mean falls between 4800 psi and 5200 psi?

(b) What sample n would be necessary in order to have P(4950 < ?̅<5050) = 0.95 (0.99)?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	5000
std deviation   =σ=	400.000
sample size       =n=	25
std error=σx̅=σ/√n=	80.000

probability =P(4800<X<5200)=P((4800-5000)/80)<Z<(5200-5000)/80)=P(-2.5<Z<2.5)=0.9938-0.0062=0.9876

b)

for 95% confidence interval:

	for95% CI crtiical Z          =	1.960
	standard deviation σ=	400.000
	margin of error E =	50
required sample size n=(zσ/E)2                  =		246

  for 99% confidence interval:

	for99% CI crtiical Z          =	2.576
	standard deviation σ=	400.000
	margin of error E =	50
required sample size n=(zσ/E)2                  =		425