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A summer resort rents rowboats to customers but does not allow more than four people to a boat. Each boat is designed to hold no more than 800 pounds. Suppose the distribution of adult males who rent boats, including their clothes and gear, is normal with a mean of 190 pounds and standard deviation of 10 pounds. If the weights of individual passengers are independent, what is the probability that a group of four adult male passengers will exceed the acceptable weight limit of 800 pounds?

Respuesta :

Answer:

.023

Explanation:

190 x 4 = 760

New SD is 10 squared x 4 (people) and sqrt all that down to 20

Draw yourself a normal model. What's the P(the combined weights are at least 800 or above) This is an upper tail test.

So you find your z score (if it weren't normal than a t-score for means).

800-760/20 = 2

On your calc... under 2nd vars.....normcdf(2,9^9)

Which gives you .02275 rounded to .023 and that's A for you.

The probability of a group of 4 adult male passengers exceeding the acceptable weight restriction of 800 pounds would be:

- 0.023

Given that,

The maximum number of people to be present in the boat = 4

The maximum weight it can hold = 800 pounds

Mean = 190 pounds

Standard Deviation = 10 pounds

To find,

Probability of a group of 4 adult male passengers exceeding the acceptable weight restriction of 800 pounds = ?

P(∑X ≥ 800) [tex]=[/tex] [tex]1 - P[/tex]((∑X < 800)

[tex]=[/tex] [tex]1 - P[/tex]{(∑X - [tex]n[/tex]μ)/σ[tex]\sqrt{n}[/tex]} < {(800 - 760)/20)}

[tex]=[/tex] [tex]1 - P[/tex][tex](z < 2)[/tex]

= [tex]1 - 0.9772[/tex]

[tex]= 0.023[/tex]

Thus, 0.023 is the correct answer.

Learn more about 'Probability' here:

brainly.com/question/11234923

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