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kondor19780726 [428]
1 year ago
9

An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is

approximately normal with μ = 15.5 and σ = 3.6. What is the probability that during a given week the airline will lose less than 20 suitcases?
Mathematics
1 answer:
Viktor [21]1 year ago
4 0

Answer:

\frac{?}{?}

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JulsSmile [24]
<h3>Answer:  24416.64</h3>

Work Shown:

V = (4/3)*\pi*r^3\\\\V \approx (4/3)*3.14*18^3\\\\V \approx 24416.64\\\\

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2 years ago
Which expression is equivalent to this one?
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Hello,

Which expression is equivalent to this one?

1/6(x - 18)

Distributive property: (1/6x) - (1/6 x 18)

= 1/6x

= -18

Answer: 1/6x * (-18)

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The perimeter of a rectangular field is 322 m.
Bond [772]

Answer:

75

Step-by-step explanation:

322 = 2(86) + 2w

322 = 172 + 2w

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5 0
2 years ago
For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 50
Volgvan

Answer:

P(\bar X>530)=1-0.808=0.192

Step-by-step explanation:

1) 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".  

Let X the random variable that represent the scores, and for this case we know the distribution for X is given by:

X \sim N(\mu=505,\sigma=170)  

And let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(505,\frac{170}{\sqrt{35}})

2) Calculate the probability

We want this probability:

P(\bar X>530)=1-P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >530)=1-P(Z

P(\bar X>530)=1-0.808=0.192

3 0
3 years ago
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