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the baggage weights for passengers using a domestic airline are normally distributed with a mean of 22 lbs. and a standard deviation of 4 lbs. assume that the limit on total luggage weight is 2250 lbs. if 100 passengers are aboard the airline, what is the probability that their total baggage weight exceeds the limit?

Respuesta :

The probability that the total baggage weight exceeds the limit is 0.1056.

The normal distribution is a continuous probability distribution with most values located close to the center peak and is symmetrical around its mean.

For example,  heights, measurement errors, blood pressure, and IQ scores are normally distributed because the majority of people have these values close to the standard value.

Here, the mean [tex]\mu[/tex] is 22 lbs, the standard deviation [tex]\sigma[/tex] is 4 lbs, and the sample size n is 100.

The total baggage weight of the passengers is represented as [tex]\sum x[/tex].

Then,

[tex]\begin{aligned} P\left(\sum x > 2250\right) &= P\left(\frac{\sum x}{ n} > \frac{2250}{100}\right)\\& = P( \bar{X} > 22.5) \end{aligned}[/tex]

We know, that [tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

So, representing in a normal distribution,

[tex]\begin{aligned}P\left[z > \frac{(22.5 - 22)}{\frac{4}{\sqrt100}}\right] &= P(z > 1.25) \\&= 0.1056\end{aligned}[/tex]

From the z-distribution table, the value for z>1.25 is 0.1056.

Therefore, the answer is 0.1056.

To know more about normal distribution:

https://brainly.com/question/15103234

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