An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable x with a mean value of 49 lb and a standard deviation of 19 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With n = 100, the total weight exceeds the limit when the average weight x exceeds 6000/100.) (Round your answer to four decimal places.)

Accepted Solution

Answer:The approximate probability that the total weight of their baggage[tex]P (\bar x >60) = 0.000[/tex]Step-by-step explanation:Given data;number of passenger is 100baggage limit = 6000 lbstandard deviation = 19 lbmean value  = 49For 100 passengers baggage limit is 6000 lbso, average weight for per passenger  > 60[tex]P (\bar x >60)[/tex]As mean value is 49, therefore  [tex]P (\bar x >60)[/tex] lie on right side centerz is given as[tex]z = \frac{ \bar x \mu}{\sigma_{\bar x}}[/tex][tex]z = \frac{60 - 49}{\frac{\sigma}{\sqrt n}}[/tex][tex]z  = \frac{60 - 49}{\frac{19}{\sqrt {100}}} = 5.79[/tex][tex]P(Z>5.79) = AREA to the right of 5.79[/tex][tex]P (\bar x>60)  = P (Z>5.79)  = 1 -P (Z < 5.79)[/tex]= 1 -P (Z < 5.79) = 1 - 1.0000= 0.000 [tex]P (\bar x >60) = 0.000[/tex]