Question

Given that 10% of the nails made using a certain manufacturing process have a length less than 2.48 inches, while 5% have a length greater than 2.54 inches, what are the mean and standard deviation of the lengths of the nails? Assume that the lengths have a normal distribution.

Answer 1

Answer:

The mean, μ is 2.5063 and

The standard deviation, σ = 2.0499 × 10⁻²

Step-by-step explanation:

Here we have

$z = \frac{x- \mu}{\sigma}$

$z \sigma ={x- \mu}{}$ and

10% have length < 2.48 in while

5% have length > 2.54 in

and

From the z score table

Critical z at 10% to the left = -1.282

Critical z at 5% to the right = 1.645

Therefore, the equations become

-1.282σ = 2.48 - μ → μ -1.282σ = 2.48

1.645σ = 2.54 - μ → μ + 1.645σ = 2.54

Solving the above equations gives

The mean, μ = 2.5063 and

The standard deviation, σ = 2.0499 × 10⁻².