Question

1. A company supplies pins to a customer. It uses an automatic lathe to produce the pins. Due to factors such as vibration, temperature and wear and tear, the lengths of the pins and normally distributed with a mean of 25.30 mm and a standard deviation of 0.45 mm. The customer will only buy pins with lengths in the interval 25.00 ± 0.50 mm.

Answer 1

The percentage of the pins that will be acceptable to the customer is 63.16%.

What will the percentage be?

Based on the information, the probabilty that the pin lies is between 24.5 and 25.5. This will be illustrated as P(24.5 <x < 25.5).

So convert this into Z score, will be:

P(24.5 <x < 25.5):

= P((24.5-25.3)/0.45 <Z < (25.5-25.3)/0.45)

Solving this, we will get

P(-16/9<Z < 4/9) = P(-1.77 <Z < 0.44),

By looking at the z table and solving for Z by using P (Z<0.44)-P(Z>-1.77) will be:

= 0.67 -(1-0.9616)

= 0.6316

= 63.16%

Learn more about probability on:

brainly.com/question/24756209

#SPJ1

Complete question

A company supplies pins to a customer. It uses an automatic lathe to produce the pins. Due to factors such as vibration, temperature and wear and tear, the lengths of the pins and normally distributed with a mean of 25.30 mm and a standard deviation of 0.45 mm. The customer will only buy pins with lengths in the interval 25.00 ± 0.50 mm. What percentage of the pins will be acceptable to the customer?