Question

Find the upper 20%of the weight?

Answer 1

Answer:

The upper 20% of the weighs are weights of at least X, which is $X = 0.84\sigma + \mu$, in which $\sigma$ is the standard deviation of all weights and $\mu$ is the mean.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Upper 20% of weights:

The upper 20% of the weighs are weighs of at least X, which is found when Z has a p-value of 0.8. So X when Z = 0.84. Then

$Z = \frac{X - \mu}{\sigma}$

$0.84 = \frac{X - \mu}{\sigma}$

$X = 0.84\sigma + \mu$

The upper 20% of the weighs are weights of at least X, which is $X = 0.84\sigma + \mu$, in which $\sigma$ is the standard deviation of all weights and $\mu$ is the mean.