Question

The lengths of nails produced in a factory are normally distributed with a mean of 3.34 centimeters and a standard deviation of 0.07 centimeters. Find the two lengths that separate the top 3% and the bottom 3%. These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.

Answer 1

Answer:

3.47 and 3.21

Step-by-step explanation:

Let us assume the nails length be X

$X \sim N(3.34,0.07^2)$

Value let separated the top 3% is T and for bottom it would be B

$P(X < T)= 0.97$

Now converting, we get

$P(Z < \frac{T-3.34}{0.07})= 0.97$

Based on the normal standard tables, we get

$P(Z < 1.881)= 0.97$

Now compare these two above equations

$\frac{T-3.34}{0.07} = 1.881 \\\\ T = 1.881 \times 0.07 + 3.34 \\\\ = 3.47$

So for top 3% it is 3.47

Now for bottom we applied the same method as shown above

$P(Z < \frac{B-3.34}{0.07})= 0.03$

Based on the normal standard tables, we get

$P(Z < -1.881)= 0.03$

Now compare these two above equations

$\frac{B-3.34}{0.07} = -1.881$

$= -1.881 \times 0.07 + 3.34 \\\\ = 3.21$

hence, for bottom it would be 3.21