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Sidana [21]
3 years ago
15

Amy, Bess, Cassie, Diana, and Ellen are a basketball team. Two of them will be guards. Assuming they all have an equal chance of

becoming guards, what is the probability that Diana will be a guards. which outcome (or outcomes) of the sample space compose the event? Describe the probability Diana becoming a guard as impossiable, unlikey, neither likey nor unlikey, likey or certain. Justify your answ
Mathematics
2 answers:
taurus [48]3 years ago
6 0

Answer: 1: The sample space for this problem is {Amy, Bess, Cassie, Diana, Ellen} 2: 10 outcomes 3: 2/5 or 0.4 4: The probability of Diana becoming a guard is possible since her chance of being chosen is 40%.

Step-by-step explanation: The explanations are on every problem except for 2 so here it is. The outcomes of the sample problem are Diana and Ellen, Diana and Cassie, Diana and Amy, Diana and Bess, Amy and Bess, Amy and Cassie, Amy and Ellen, Cassie and Bess, Cassie and Ellen, Ellen and Bess which adds up to ten outcomes since two people can be chosen from the group of five.

Alenkasestr [34]3 years ago
4 0
There are 5 people who can be guards, 2 will be guards. Since each person has the same chance of becoming a guard, each person has a 2/5 chance of being a guard which makes it unlikely for Diana to become a guard.
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Ainat [17]

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99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

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

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 height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

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.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

  P(\bar X \geq 65.86 inches) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } } ) = P(Z \geq -2.57) = P(Z \leq 2.57)

                                                                        = <u>0.99492  or  99.5%</u>

<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

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Answer:

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Step-by-step explanation:

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