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

Answer:

$0.2564\text{ pounds}$

Step-by-step 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 $5.1282-4.8718=\boxed{0.2564\text{ pounds}}$.

Answer 2

Answer:

0.2564

Step-by-step 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