A machine produces 1000 steel O-rings per day. Each ring has probability 0.9 of meeting a thickness specification.

a. What is the probability that on a given day, fewer than 890 O-rings meet the specification?

b. Find the 60th percentile of the number of O-rings that meet the specification.

c. If the machine operates for five days, what is the probability that fewer than 890 O-rings meet the specification on three or more of those days?

Step 1 of 4</p>

Here given a machine produces 1000 steel O-rings per day

n=1000

The probability of each ring meeting the specification is p=0.9

Let x represents the no.of rings meeting the specification

Here XB(1000,0.9)

Here x is approximately following the normal distribution

Mean =np

=1000(0.9)

=900

Variance

=1000(0.9)(0.1)

=90

SD

=9.49

Step 2 of 4</p>

a) Here we have to find the probability of fewer than 890 O rings meet the specification

We have to find P(X<890)

Now Z=(

=(890-900)/9.49

=-1.05

Now find P(Z< -1.05) value from the standard normal tables

P(Z< -1.05)=0.1469

The probability of fewer than 890 O rings meet the specification is 0.1469