Question

The number of nails of a given length is normally distributed with a mean length of 5 in. and a standard deviation of 0.03 in. In a bag containing 120 nails, how many nails are between 4.97 in. long and 5.03 in. long?

Answer 1

[AI-generated answer — Bot GPT 3.5]

To solve this problem, we can use the properties of the normal distribution.

First, let's find the z-scores for the lower and upper limits. The z-score formula is:

z = (x - μ) / σ

where x is the observed value, μ is the mean, and σ is the standard deviation.

For the lower limit:
z1 = (4.97 - 5) / 0.03 = -1

For the upper limit:
z2 = (5.03 - 5) / 0.03 = 1

Next, we can use a standard normal distribution table or a calculator to find the probability of being within these two z-scores.

From the standard normal distribution table, we can find that the probability of being within -1 and 1 is approximately 0.6827.

So, the probability of a nail being between 4.97 in. and 5.03 in. long is 0.6827.

To find the absolute number of nails, we multiply this probability by the total number of nails in the bag:

Number of nails = probability * total number of nails
= 0.6827 * 120
≈ 81.92

Rounding to the nearest whole number, we can conclude that there are roughly 82 nails between 4.97 in. long and 5.03 in. long in the bag.