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drek231 [11]
4 years ago
8

Suppose the chance of winning a game is 1/50. If you play the game 50 times independently, is it certain that you will win? If n

ot, what is the chance of winning at least once?
a.0.0200
b.0.9462
c.0.6358
d.1.0000
Mathematics
1 answer:
arlik [135]4 years ago
5 0

We are given chance of winning a game is 1/50.

In decimal form 1/50 could be written as 0.02.

So, for each game, the chance of winning is 0.02.

Therefore, chance of loss would be = 1 -(0.02) = 0.98.

Chance of loss of 50 independent games = 0.98*0.98*0.98*0.98...... upto 50 times, that is (0.98)^50.

The chance of winning at least once = 1 - (0.98)^50.

We got (0.98)^50 = 0.36417

Therefore, 1 - (0.98)^50 = 1 - 0.36417.

We get 0.63583.

Therefore, correct option is c. 0.6358.

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Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random
Fudgin [204]

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Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

<em>Let </em>\bar X<em> = sample mean</em>

The z-score probability distribution for sample mean is given by;

               Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean = $50,000

            \sigma = population standard deviation = $25,000

            n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P(\bar X > $52,000)

  P(\bar X > $52,000) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } } ) = P(Z > 2.53) = 1 - P(Z \leq 2.53)

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<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

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What is 3.30 divided by -2.00? Show work.
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Answer:

<h2>1.65</h2>

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\frac{3.3}{-2}\\\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{-b}=-\frac{a}{b}\\=-\frac{3.3}{2}\\\\\mathrm{Divide\:the\:numbers:}\:\\\frac{3.3}{2}=1.65

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3 years ago
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