Question

An industrial sewing machine uses ball bearings that are targeted to have a diameter of

0.73

inch. The lower and upper specification limits under which the ball bearing can operate are

0.72

inch​ (lower) and

0.74

inch​ (upper). Past experience has indicated that the actual diameter of the ball bearings is approximately normally​ distributed, with a mean of

0.732

inch and a standard deviation of

0.006

inch. Suppose a random sample of

21

ball bearings are selected. Complete parts​ (a) through​ (e).

a. What is the probability that the sample mean is between the target and the population mean of

0.732​?

b. What is the probability that the sample mean is between the lower specification limit and the​ target?

c. What is the probability that the sample mean is greater than the upper specification​ limit?

d. What is the probability that the sample mean is less than the lower specification​ limit?

Answer 1

for normal distribution z score =(X-μ)/σx
mean μ=	0.732
standard deviation σ=	0.006

std error=σx̅=σ/√n=

0.0013

a)

probability =P(0.73<X<0.732)=P((0.73-0.732)/0.001)<Z<(0.732-0.732)/0.001)=P(-1.53<Z<0)=0.5-0.063=0.4370

b)

probability =P(0.72<X<0.74)=P((0.72-0.732)/0.001)<Z<(0.74-0.732)/0.001)=P(-9.17<Z<6.11)=1-0=1.0000

c)

probability =P(X>0.74)=P(Z>(0.74-0.732)/0.001)=P(Z>6.11)=1-P(Z<6.11)=1-1=0.0000

d)

probability =P(X<0.72)=(Z<(0.72-0.732)/0.001)=P(Z<-9.17)=0.0000