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Anna [14]
3 years ago
7

In Evans history class, 10 out of 100 key terms will be randomly selected to appear on the final exam; Evan must then choose 7 o

f those 10 to define. Since he knows the format of the exam in advance, Evan is trying to decide how many key terms he should study. (a) Suppose that Evan decides to study s key terms, where s is an integer between 0 and 100. Let X be the number of key terms appearing on the exam that he has studied. What is the distribution of X? Give the name and parameters, in terms of s. (b) Calculate the probability that Evan knows at least 7 of the 10 key terms that appear on the exam, assuming that he studies s = 75 key term
Mathematics
1 answer:
motikmotik3 years ago
6 0

Answer:

(a) Hypergeometric distribution

(b) 0.785384...

Step-by-step explanation:

(a) looking at the question, we see that X = \in \{0, ... , 10 \}, suppose Evans studied s key terms, this implies that he has not studied 100 - s key terms. Suppose k = \in \{0, ... , 10 \} if X = k, then k out of the s key terms he had studied appeared. But 10 - k out of 100 - s key terms he hasn't studied appeared. Thus X is an Hypergeometric distribution, X~HGeom(s, 100 - s, 10) with PMF:

P_{x}(k) = P(X = k)

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Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be X centimeters.

X follows a normal distribution with

  • mean \mu = 11.2 and
  • standard deviation \sigma = 2.1.

In other words,

X\sim \text{N}(11.2, 2.1^{2}),

and the probability in question is X \le 13.

Z-score table approach:

Find the z-score of this measurement:

\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143. Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that Z will be less than or equal to the z-score in question. (In case the question is asking for the probability that Z is greater than the z-score, subtract the value from table from 1.)

P(X\le 13) = P(Z \le 0.857143) \approx 0.8051.

"Technology" Approach

Depending on the manufacturer, the steps generally include:

  • Locate the cumulative probability function (cdf) for normal distributions.
  • Enter the lower and upper bound. The lower bound shall be a very negative number such as -10^{9}. For the upper bound, enter 13
  • Enter the mean and standard deviation (or variance if required).
  • Evaluate.

For example, on a Texas Instruments TI-84, evaluating \text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 ) gives 0.804317.

As a result,

P(X\le 13) = 0.804317.

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

  • Number of trials n: 150.
  • Probability of success p: 0.804317.

Let Y be the number of green beans out of this 150 that are shorter than 13 centimeters. Y\sim\text{B}(150,0.804317).

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

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