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frez [133]
3 years ago
5

Suppose x is a uniform random variable with values ranging from 40 to 90. Find the probability that a randomly selected observat

ion exceeds 75.
Mathematics
1 answer:
maxonik [38]3 years ago
6 0

ok, so x is a number ranging from 40-90 which is 50 meaning that x has 50 different possibilities. But what are the chances that x is 75? well the chances of x being 75 is 1 out of 50.

so that means there is a 2% chance of x being 75

brainliest is appreciated

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Numerator is the the number above the line on a fraction, while the denominator is the number underneath the line. 

\frac{6}{-8} is similar to \frac{-6}{8}

Fraction with numerator -6 and denominator 8 
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3 years ago
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Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election. Assume the probab
Liono4ka [1.6K]

Answer:

0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

153 voters:

This means that n = 153

Assume the probability that a given registered voter will vote in the presidential election is 63%.

This means that p = 0.63

Mean and standard deviation:

\mu = E(X) = np = 153(0.63) = 96.39

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{153*0.63*0.37} = 5.97

Consider the probability that at least 88 out of 153 registered voters will vote in the presidential election.

Using continuity correction, this is: P(X \geq 88 - 0.5) = P(X \geq 87.5), which is 1 subtracted by the p-value of Z when X = 87.5.

Z = \frac{X - \mu}{\sigma}

Z = \frac{87.5 - 96.39}{5.97}

Z = -1.49

Z = -1.49 has a p-value of 0.0681.

1 - 0.0681 = 0.9319

0.9319 = 93.19% probability that at least 88 out of 153 registered voters will vote in the presidential election.

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3 years ago
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2 years ago
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