Question

A data set lists weights​ (lb) of plastic discarded by households. The highest weight is 5.24 ​lb, the mean of all of the weights is x overbarequals1.932 ​lb, and the standard deviation of the weights is sequals1.135 lb. a. What is the difference between the weight of 5.24 lb and the mean of the​ weights? b. How many standard deviations is that​ [the difference found in part​ (a)]? c. Convert the weight of 5.24 lb to a z score. d. If we consider weights that convert to z scores between minus2 and 2 to be neither significantly low nor significantly​ high, is the weight of 5.24 lb​ significant?

Answer 1

Answer:

a) 3.308 lb

b) That difference is 2.91 standard deviations.

c) $Z = 2.91$

d) We have that $Z = 2.91 > 2$, so the weight of 5.24lb is significantly high.

Step-by-step explanation:

The z-score formula is given by:

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

In which: X is the weight we are going to find the z-score of, $\mu$ is the mean of all the weights and $\sigma$ is the standard deviation of all the weights.

The problems gives these following informations:

The highest weight is 5.24 ​lb

The mean of all the weights is 1.932 lb

The standard deviation of all the weights is 1.135 lb

a. What is the difference between the weight of 5.24 lb and the mean of the​ weights?

$D = 5.24lb - 1.932lb = 3.308lb$

b. How many standard deviations is that​ [the difference found in part​ (a)]?

The difference found in (a) is 3.308lb.

This can be calculated by a simple rule of three.

1 standard deviation - 1.135lb

x standard deviations - 3.308lb

$1.135x = 3.308$

$x = \frac{3.308}{1.135}$

$x = 2.91$

That difference is 2.91 standard deviations.

c. Convert the weight of 5.24 lb to a z score.

We have that:

$X = 5.24, \mu = 1.932$ and $\sigma = 1.135$. So:

$Z = \frac{X - \mu}{\sigma} = \frac{5.24 - 1.932}{1.135} = 2.91$