Question

The average weight of an airline passenger’s suitcases is 45 pounds with a standard deviation of 2 pounds. If 16% of the suitcases are overweight, find the maximum weight allowed by the airline. Assume the variable is bell – shaped distributed. Hint: Sketch and label a normal model to help answer the question.

Answer 1

Answer:

we over weighted suitcase will be 46.9889 pound

Step-by-step explanation:

We have given the average weight of passenger's suitcase $\mu =45\ pounds$

Standard deviation $\sigma =2\ pounds$

16 % of the suitcase are overweight so 100-16 =84 % of the suitcase will pass without any problem

So probability $p=0.84$

From z table at p =0.84 z will be z=0.99445

We know that $z=\frac{x-\mu }{\sigma }$

$0.99445=\frac{x-45}{2}$

x = 46.9889 pound

So we over weighted suitcase will be 46.9889 pound

Answer 2

Using standard deviation formula to solve the problem. Then the overweighted suitcase is 46.9889 pounds.

What is a standard deviation?

It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.

Given

The average $(\mu )$ weight of an airline passenger’s suitcases is 45 pounds.

The standard deviation $(\sigma)$ of 2 pounds.

16% of the suitcase are overweight. So,

$100 - 16 = 84\%$

The suitcase will pass without any problem. So the probability will be

$\rm P = 0.84$

From z table at P = 0.84 will be

z = 0.99445

We know that

$\begin{aligned} z &= \dfrac{x - \mu}{\sigma}\\0.99445 &= \dfrac{x - 45}{2}\\x &= 46.9889\end{aligned}$

Thus, the overweighted suitcase is 46.9889 pounds.

More about the standard deviation link is given below.

https://brainly.com/question/12402189