Question

In a survey, three out of four students said that courts show too much concern for criminals. Of seventy randomly selected students. Would 68 be considered an unusually high number of students who feel that courts are partial to criminalsA. The correct answer is not among the choices.
B. Yes, because P(X ≥ 68) ≤ .05
C. Yes, because P(X ≤ 68) ≤ .05
D. No,because P(X≥68)≤.05
E. No,because P(X≤68)≤.05

Answer 1

Answer:

B. Yes, because P(X ≥ 68) ≤ .05

Step-by-step explanation:

The probability of a score X occuring is said to be unusually high if the probability of a score of X or larger is lower than 5%.

For each student, there are only two possible outcomes. Either they think that the court shows too much concern for criminals, or they do not think this. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

Three out of four students said that courts show too much concern for criminals. This means that $p = \frac{3}{4} = 0.75$

Would 68 be considered an unusually high number of students who feel that courts are partial to criminals?

We have to find $P(X \geq 68)$.

$P(X \geq 68) = P(X = 68) + P(X = 69) + P(X = 70)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 68) = C_{70,68}.(0.75)^{68}.(0.25)^{2} < 0.000001$

$P(X = 69) = C_{70,69}.(0.75)^{69}.(0.25)^{1} < 0.000001$

$P(X = 70) = C_{70,70}.(0.75)^{70}.(0.25)^{0} < 0.000001$

So

$P(X \geq 68) = P(X = 68) + P(X = 69) + P(X = 70) < 0.000003$

$P(X \geq 68)$ is lower than 0.05, so this is an unusually high number.

So the correct answer is:

B. Yes, because P(X ≥ 68) ≤ .05