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Bas_tet [7]
3 years ago
11

What is not a potential source of bias: Of the 2000 students attending, every 5th student going into the basketball game is aske

d if the school should spend more money on sports. Question 8 options: Students at the game are more likely to be interested in sports. Asking only students, not adults. Not a large enough sample. The survey is conducted at only one sport event.
Mathematics
2 answers:
ahrayia [7]3 years ago
8 0
I think it is C, not a large enough sample.
Nezavi [6.7K]3 years ago
7 0

Answer:

Not a large enough sample

Step-by-step explanation:

Every 5th student going into the basketball game implies that 1/5 of the attendance would be surveyed, that make it a representative sample, so it is not a potential source of bias.

A biased sample implies that the sample doesn't represent the population. If you ask people if the school should spend more money on sports in a sport event, they are likely to say yes.

If you ask only students, not adults, you are avoiding people in purpose.

To get a representative sample, you should consider more than one sport.

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Step-by-step explanation:

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Answer:

(a) The value of <em>x</em> is 0.30.

(b) The probability that a reported case of rabies is not a raccoon is 0.45.

(c) The probability that a reported case of rabies is either a bat or a fox is 0.15.

Step-by-step explanation:

Denote the events as follows:

<em>R</em> = reported case of rabies is a raccoon

<em>F</em> = reported case of rabies is a fox

<em>B</em> = reported case of rabies is a bat

<em>O</em> = reported case of rabies is a some other animal.

The data provided is:

P (R) = 0.55

P (F) = 0.11

P (B) = 0.04

P (O) = <em>x</em>.

(a)

A property of a probability distribution is that the sum of all individual properties is 1.

That is, \sum P(X)=1

Compute the value of <em>x</em> as follows:

\sum P(X)=1\\P(R)+P(F)+P(B)+P(O)=1\\0.55+0.11+0.04+x=1\\0.70+x=1\\x=1-0.70\\x=0.30

Thus, the value of <em>x</em> is 0.30.

(b)

The probability of the complement of an event is the probability of its not happening.

P(A^{c})=1-P(A)

Compute the probability that a reported case of rabies is not a raccoon as follows:

P(R^{c})=1-P(R)\\=1-0.55\\=0.45

Thus, the probability that a reported case of rabies is not a raccoon is 0.45.

(c)

Compute the probability that a reported case of rabies is either a bat or a fox as follows:

P(B\cup F)=P(B)+P(F)\\=0.11+0.04\\=0.15

Thus, the probability that a reported case of rabies is either a bat or a fox is 0.15.

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Answer:

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Step-by-step explanation:

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