Question

A company manufactures televisions. The average weight of the televisions is 5 pounds with a standard deviation of 0.1 pound. Assuming that the weights are normally distributed, what is the weight that separates the bottom 10% of weights from the top 90%?​

Answer 1

0.2564

Explanation:

90th percentile, we use the formula X=μ + Zσ,

Where u = mean and sigma = standard deviation and Z = 1.282

The mean is 5 and sigma = .1

X = 5+1.282(.1)

X = 5.1282

10th percentile, we use the formula X=μ + Zσ,

Where u = mean and sigma = standard deviation and Z = -1.282

The mean is 5 and sigma = .1

X = 5-1.282(.1)

X = 4.8718

The difference is

5.1282 - 4.8718

0.2564

Answer 2

`0.2564\text{ pounds}`

Explanation:

The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.

To find the
`X` percentile for the television weights, use the formula:

`X=\mu +k\sigma`, where
`\mu` is the average of the set,
`k` is some constant relevant to the percentile you're finding, and
`\sigma` is one standard deviation.

As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute
`\mu=5`,
`k=1.282`, and
`\sigma=0.1`:

`X=5+(1.282)(0.1)=5.1282`

Therefore, the 90th percentile weight is 5.1282 pounds.

Repeat the process for calculating the 10th percentile weight:

`X=5+(-1.282)(0.1)=4.8718`

The difference between these two weights is
`5.1282-4.8718=\boxed{0.2564\text{ pounds}}`.