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-BARSIC- [3]
3 years ago
7

The collection of all possible outcomes of a random phenomenon is called A. the sample space. B. the distribution. C. the probab

ility. D. a census. E. None of the above.
Mathematics
1 answer:
Hoochie [10]3 years ago
8 0

Answer:

Option A is correct.

Sample space is the collection of all possible outcomes of a random phenomenon.

Step-by-step explanation:

- Sample Space

The sample space of a phenomenon is the set of all possible outcomes for that phenomenon.

- Distribution

The distribution of a statistical data set is defined as a listing or function showing all the possible values (or intervals) of the data and how often they occur.

- Probability

Probability describes how likely something is to happen. Probability values are usually between 0 and 1; with 0 denoting impossibility and 1 denoting total certainty.

- Census

Census is an official count or survey of a population

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A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for their most recent
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Answer:

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(b) P (0.15 < p' < 0.20) = 0.7784.

(c) P (133 < X < 171) = 0.2205

Step-by-step explanation:

Let <em>p</em> = proportion of business trip that were for  internal company visit.

It is provided that 19% responded that it was for an internal company visit, i.e.

<em>p</em> = 0.19.

A sample of <em>n</em> = 950 business travelers are randomly selected.

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

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 \mu_{\hat p}=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

Thus, the distribution of \hat p is N (<em>μ</em> = 0.19, <em>σ </em>= 0.013).

(a)

Compute the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:

P (p' > 0.25) = P (Z > 4.62) = 0

*Use a <em>z</em>-table for the probability.

Thus, the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.

(b)

Compute the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:

P (0.15 < p' < 0.20) = P (-3.08 < Z < 0.77)

                               = P (Z < 0.77) - P (Z < -3.08)

                               = 0.7794 - 0.0010

                               = 0.7784

*Use a <em>z</em>-table for the probability.

Thus, the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.7784.

(c)

Compute the proportion of <em>X</em> = 133  and <em>X</em> = 171 as follows:

p' = 133/950 = 0.14

p' = 171/950 = 0.18

Compute the value of P (0.14 < p' < 0.20) as follows:

P (0.14 < p' < 0.18) = P (-3.85< Z < -0.77)

                               = P (Z < -0.77) - P (Z < -3.85)

                               = 0.2206- 0.0001

                               = 0.2205

*Use a <em>z</em>-table for the probability.

Thus, the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.2205.

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