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Wittaler [7]
3 years ago
15

On the spinner, the probability of landing on black is 1/2 , and the probability of landing on red is 1/3. Create a probability

distribution for the experiment of spinning the spinner.
Mathematics
1 answer:
Margarita [4]3 years ago
7 0

Answer:

1) p_i \geq 0 , \forall i

2)\sum_{i=1}^n P_i = 1, i =1,2,...,n

And for this case we have:

\frac{1}{2}+\frac{1}{6}= \frac{2}{3}

By the complement rule we can find the probability that the spinner land in a non black or red space:

p(N) = 1- \frac{1}{2} -\frac{1}{3}= \frac{1}{6}

And then the probability distribution would be:

Color      Red     Black    N

Prob.      1/3         1/2       1/6

Step-by-step explanation:

For this case we have two possible outcomes for the spinner experiment:

p(black) =\frac{1}{2}

p(red) = \frac{1}{3}

In order to have a probability distribution we need to satisfy two conditions:

1) p_i \geq 0 , \forall i

2)\sum_{i=1}^n P_i = 1, i =1,2,...,n

And for this case we have:

\frac{1}{2}+\frac{1}{6}= \frac{2}{3}

By the complement rule we can find the probability that the spinner land in a non black or red space:

p(N) = 1- \frac{1}{2} -\frac{1}{3}= \frac{1}{6}

And then the probability distribution would be:

Color      Red     Black    N

Prob.      1/3         1/2       1/6

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Step-by-step explanation:

So using long division, you can solve for the quotient and the remainder.

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Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the pro
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Answer:

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

Step-by-step explanation:

Normal Probability Distribution

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

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that \mu = 0, \sigma = 1

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

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

Z = \frac{-1.69 - 0}{1}

Z = -1.69

Z = -1.69 has a p-value of 0.0455

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

Z = \frac{-2.23 - 0}{1}

Z = -2.23

Z = -2.23 has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

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