Mathematics, 06.10.2019 08:30, ashcookie27

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.)

Answers: 2

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Answers

answered: xmengeo

The approximate probability that the total weight of their baggage

$P (\bar x 60) = 0.000$

Step-by-step explanation:

Given data;

number of passenger is 100

baggage limit = 6000 lb

standard deviation = 19 lb

mean value = 49

For 100 passengers baggage limit is 6000 lb

so, average weight for per passenger > 60

$P (\bar x 60)$

As mean value is 49, therefore $P (\bar x 60)$ lie on right side center

z is given as

$z = \frac{ \bar x \mu}{\sigma_{\bar x}}$

$z = \frac{60 - 49}{\frac{\sigma}{\sqrt n}}$

$z = \frac{60 - 49}{\frac{19}{\sqrt {100}}} = 5.79$

P(Z5.79) = AREA to the right of 5.79

$P (\bar x60) = P (Z5.79) = 1 -P (Z < 5.79)$

= 1 -P (Z < 5.79)

= 1 - 1.0000

= 0.000

$P (\bar x 60) = 0.000$

answered: Guest

a 360 degree rotation

step-by-step explanation:

answered: Guest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

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An airplane with room for 100 passengers has a total baggage limit of 6000 lb. suppose that the tota...