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pav-90 [236]
3 years ago
7

You take a quiz with 6 multiple choice questions. After you studied your estimated that you would have about an 80 % chance of g

etting any individual question right what are your chances of getting them all right use at least 20 trials?
Mathematics
1 answer:
Arada [10]3 years ago
4 0

Answer:

1.15%

Step-by-step explanation:

To get the probability of m  independent events you multiply the individual probability of each event. In this case we have m independent events, each one with the same probability, therefore:

p^{m}

0.8^{20} = 1.15\%

This is a particlar scenario of binomial distribution problem. So the binomial distribution questions are about the number of success of m independent events, where every individual event has the same p probability. In the question we have 20 events and each event has a probability of 80%. The binomial distribution formula is:

\binom{n}{k} * p^{k} * (1-p)^{n-k}

n is the number of events

k is the number of success

p is the probability of each individual event

\binom{n}{k} is the binomial coefficient

the binomial coefficient allows to find the subsets of k elements in a set of n elements.  In this case there is only one subset possible since the only way to get 20 of 20 correct questions is to getting right all questions (for getting 19 of 20 questions there are many ways, for example getting the first question wrong and all the other questions right, or getting second questions wrong and all the other questions right, etc).

\binom{n}{k} = \frac{n!}{k!(n-k)!}

therefore, for this questions we have:

\frac{20!}{20!(20-20)!} * 0.8^{20} * (1-0.8)^{0} = 1.15\%

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The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room
Alexandra [31]

Answer:

\bar x = 260.1615

\sigma = 70.69

The confidence interval of standard deviation is: 53.76 to 103.25

Step-by-step explanation:

Given

n =20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

\bar x = \frac{\sum x}{n}

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}

\bar x = \frac{5203.23}{20}

\bar x = 260.1615

\bar x = 260.16

Solving (b): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}\sigma = \sqrt{\frac{94938.80}{19}}

\sigma = \sqrt{4996.78}

\sigma = 70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95

So:

\alpha = 1 -c

\alpha = 1 -0.95

\alpha = 0.05

Calculate the degree of freedom (df)

df = n -1

df = 20 -1

df = 19

Determine the critical value at row df = 19 and columns \frac{\alpha}{2} and 1 -\frac{\alpha}{2}

So, we have:

X^2_{0.025} = 32.852 ---- at \frac{\alpha}{2}

X^2_{0.975} = 8.907 --- at 1 -\frac{\alpha}{2}

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.76 to 103.25

8 0
3 years ago
The distance between Capeton and Jonesville is 80 miles. The scale on the map is 0.75 in. : 10 miles. How far apart are the citi
Mnenie [13.5K]
The answer to this question is 6 in. Hope this helps.
7 0
3 years ago
PLEASE HELPP URGENT<br> 20 points!!!
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If you're talking about the second part of the question where it asks you which facts you included in your answer, you can choose any of them and it will be counted right.
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2. The table shows the probabilities of winning or losing when the team is playing away or is playing at home.
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The total will be 08.99 because that’s how you divide and he is winning
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Find the angle of elevation of a kite on a 50m<br> string flying 40m vertically above the ground.
Lera25 [3.4K]
Sin(x)=40/50
x= sin^-1(40/50)
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angle of elevation is 53.1
6 0
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