Metal components for a machine have lengths which are normally distributed with mean 8 cm and standard deviation 4 cm.

A) the longest 10% of components require filling down. What is the smallest length of component which require filing down?

B) the shortest 15% of components have to be discarded. What is the smallest length of component which would not be discarded?

Please help

Thankyou

you can answer these and many other Z table things at

http://davidmlane.com/hyperstat/z_table.html

3(a)X=Z×sd

X=85+(1.2816)x4
X=90.1264

To find the answers to these questions, we need to use the concept of z-score. The z-score tells us how many standard deviations an individual value is from the mean.

A) To find the smallest length of component that requires filing down, we need to find the z-score that corresponds to the 10th percentile.

1. Convert the percentile to a z-score using the standard normal distribution table or a calculator. In this case, we are looking for the z-score that corresponds to the 10th percentile, which is z = -1.28 (approximately).

2. Once we have the z-score, we can use the formula: z = (x - μ) / σ, where z is the z-score, x is the value we want to find, μ is the mean, and σ is the standard deviation.

3. Rearrange the formula to solve for x: x = μ + (z * σ). Plugging in the given values, we get: x = 8 + (-1.28 * 4) = 3.88 cm.

Therefore, the smallest length of component that requires filing down is approximately 3.88 cm.

B) To find the smallest length of component that would not be discarded, we need to find the z-score that corresponds to the 85th percentile (since we want to keep the shortest 15%).

1. Convert the percentile to a z-score: The 85th percentile corresponds to z = 1.04 (approximately).

2. Use the formula: x = μ + (z * σ). Plugging in the values, we get: x = 8 + (1.04 * 4) = 12.16 cm.

Therefore, the smallest length of component that would not be discarded is approximately 12.16 cm.