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Goryan [66]
3 years ago
10

Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in

ches and a standard deviation of 3.7 inches. Fill in the blanks. In​ 95.44% of the​ years, the precipitation in this city is between​ ___ and​ ___ inches. Round your answers to the nearest tenth as needed.
Mathematics
1 answer:
nadezda [96]3 years ago
8 0

Answer:

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

Step-by-step explanation:

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".  

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

X \sim N(288,3.7)  

Where \mu=288 and \sigma=3.7

The confidence level is 95.44 and the signficance is 1-0.9544=0.0456 and the value of \alpha/2 =0.0228. And the critical value for this case is z = \pm 1.99

Using this condition we can find the limits

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

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Suppose we roll a fair six-sided die 20 times and draw ten cards from a standard 52-card deck. Let X be the number of "6"s rolle
Lera25 [3.4K]

Answer:

a) Expected value = 6.406

Variance = 4.905

Standard deviation = 2.45

b) The probability is 0.08547

Step-by-step explanation:

a) Let's suppose that:

X₁ = number of 6´s

X₂ = number of Jack, Queen, King or Aces

The mean of X₁ is:

MeanX₁ = n * p = 20 * (1/6) = 3.33

The variance of X₁ is:

Var-X_{1} =np(1-p)=3.33(1-(1/6))=2.775

The mean of X₂ is:

MeanX₂ = 10 * (16/52) = 3.076

The variance of X₂ is:

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The expect value of X is:

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The standard deviation is:

Xdevi = 4.905/2 = 2.45

b) The probability of drawing at least five six out of 20 rolls is equal to:

∑(1/6)ˣ(5/6)²⁰⁻ˣ = 0.231 with x = 5

The probability of at least 4 Jack, Queen, Kings or Aces is:

∑(16/52)ˣ(1-(16/52))¹⁰⁻ˣ = 0.37 with x = 4

The probability of given event is equal to:

P = 0.231 * 0.37 = 0.08547

5 0
3 years ago
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