Question

The length of aluminum rods produced by a company are approximated by a Gaussian distribution with a mean of 10 cm and a standard deviation of 0.02 cm. If a component is chosen at random a) what is the probability that the length of this component is between 9.98 and 10.02 cm? b) what is the probability that the length of this component is between 9.90 and 10.1 cm?

Answer 1

Given Information:

Mean length of aluminum rods = μ = 10 cm

Standard deviation of length of aluminum rods = σ = 0.02 cm

Required Information:

a) P(9.98 < X < 10.02) = ?

b) P(9.90 < X < 10.1) = ?

Answer:

a) P(9.98 < X < 10.02) = 68.27%

b) P(9.90 < X < 10.1) = 100%

Explanation:

What is Normal Distribution?

Normal Distribution or also known as Gaussian Distribution, is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability

a) We want to find out the probability that the length of aluminum rods is between 9.98 and 10.02 cm.

$P(9.98 < X < 10.02) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.98 < X < 10.02) = P( \frac{9.98- 10}{0.02} < Z < \frac{10.02 - 10}{0.02} )\\\\P(9.98 < X < 10.02) = P( \frac{-0.02}{0.02} < Z < \frac{0.02}{0.02} )\\\\P(9.98 < X < 10.02) = P( -1 < Z < 1 )$

The z-score corresponding to -1 is 0.15866 and 1 is 0.84134

$P(9.98 < X < 10.02) = P( Z < 1 ) - P( Z < -1 ) \\\\P(9.98 < X < 10.02) = 0.84134 - 0.15866 \\\\P(9.98 < X < 10.02) = 0.6827\\\\P(9.98 < X < 10.02) = 68.27 \%$

Therefore, the probability that the length of aluminum rods is between 9.98 and 10.02 cm is 68.27%

b) We want to find out the probability that the length of aluminum rods is between 9.90 and 10.1 cm.

$P(9.90 < X < 10.1) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.90 < X < 10.1) = P( \frac{9.90- 10}{0.02} < Z < \frac{10.1 - 10}{0.02} )\\\\P(9.90 < X < 10.1) = P( \frac{-0.1}{0.02} < Z < \frac{0.1}{0.02} )\\\\P(9.90 < X < 10.1) = P( -5 < Z < 5 )$

The z-score corresponding to -5 is 0 and 5 is 1

$P(9.90 < X < 10.1) = P( Z < 5 ) - P( Z < -5 ) \\\\P(9.90 < X < 10.1) = 1 - 0 \\\\P(9.90 < X < 10.1) = 1\\\\P(9.90 < X < 10.1) = 100 \%$

Therefore, the probability that the length of aluminum rods is between 9.90 and 10.1 cm is 100%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.