There are 8 runners in a race. What is the probability that Abby, Becca, and Carmen place first, second, and third in any order?

2 answers:

8 0

Well there are equally 8 runners in the race , and abby and becca and carmen are three of them . now the chances of them getting first , second , and third would be 3 out of 8 or 3 over 8 .

Mkey [24]2 years ago

3 0

$|\Omega|=8!\\
|A|=3!\cdot5!\\\\
P(A)=\dfrac{3!5!}{8!}=\dfrac{2\cdot3}{6\cdot7\cdot8}=\dfrac{1}{56}\approx1,8\%$

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emmasim [6.3K]

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2 years ago

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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil

MArishka [77]

Answer:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

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".

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,0.3)$

Where $\mu=12$ and $\sigma=0.3$

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}$