Question

1) A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 183.4-cm and a standard deviation of 1.3-cm.
Find the probability that the length of a randomly selected steel rod is between 179.9-cm and 180.3-cm.
P(179.9<x<180.3)=P(179.9<x<180.3)=

2) A manufacturer knows that their items have a normally distributed length, with a mean of 6.3 inches, and standard deviation of 0.6 inches.
If 9 items are chosen at random, what is the probability that their mean length is less than 5.9 inches?

Answer 1

1)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	183.4
std deviation   =σ=	1.3

probability that the length of a randomly selected steel rod is between 179.9-cm and 180.3-cm :

probability =P(179.9<X<180.3)=P((179.9-183.4)/1.3)<Z<(180.3-183.4)/1.3)=P(-2.69<Z<-2.38)=0.0087-0.0036=0.0051

2)

here mean=       μ=	6.3
std deviation   =σ=	0.60
sample size       =n=	9
std error=σx̅=σ/√n=	0.20000

probability =P(X<5.9)=(Z<(5.9-6.3)/0.2)=P(Z<-2)=0.0228