Question

Steel rods are manufactured with a mean length of 25 centimeters (cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed, with a standard deviation of 0.07 cm. Any rods that are shorter than 24.85 cm or longer than 25.15 cm are discarded. What proportion of rods will be discarded

Answer 1

The proportion of rods will be discarded is 0.03236

Explanation:

From the question;

mean; μ = 25 cm

Standard deviation; σ = 0.07 cm)

Now, this is a normal distribution problem where we are supposed to find the Z-score.

Z is given by;

Z = (X - μ)/0.07

Applying this to the question of, rods that are shorter than 24.85 cm or longer than 25.15 cm are discarded. We arrive at;

P(X<24.85) + P(X>25.15)

This will give us;

P(z < [(24.85 - 25)/0.07)]) + P(z > [(25.15 - 25)/0.07)])

= P(z < (-2.143)) + P(z > 2.14)

This now transforms to;

= P(z ≤ (-2.143)) + 1 - P(z ≤ 2.143)

From standard normal table which i have attached,

P(z ≤ (-2.143)) = 0.01618 and

P(z ≤ 2.143) = 0.98382

Thus,

P(X<24.85) + P(X>25.15) = 0.01618 + 1 - 0.98382 = 0.03236