Question

Suppose human weights are normally distributed with mean 175 and standard deviation 36 pounds. A helicopter is evacuating people from a building surrounded by zombies, and this helicopter can fit 9 people and with a maximum weight of 1800 pounds, i.e., an average of 200 pounds per person. If 9 people are randomly chosen to be loaded on this helicopter, what is the probability that it can safely lift off (i.e., the average weight of a person in the helicopter is less than 200)?

Answer 1

`P(\bar X <200)=P(Z<(200-175)/((36)/(√(9)))=2.083)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<2.083)=0.981`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

`X \sim N(175,36)`

Where
`\mu=175` and
`\sigma=36`

They select a sample size of n=9 people.The distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And we want to find this probability:

`P(\bar X <200)`

In order to the helicopter can safely lift off. We can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

`P(\bar X <200)=P(Z<(200-175)/((36)/(√(9)))=2.083)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<2.083)=0.981`