The following table describes the results of roadworthiness tests of Ford Focus cars that are three years old (based on data from the Department of Transportation). The random variable x represents the number of cars
That failed among six that were tested for roadworthiness: $$\begin{array} { | c | c | }
\hline x & P ( x ) \
\hline 0 & 0.377 \
\hline 1 & 0.399 \
\hline 2 & 0.176 \
\hline 3 & 0.041 \
\hline 4 & 0.005 \
\hline 5 & 0 + \
\hline 6 & 0 + \
\hline
\end{array}$$ Is the probability of getting three or more cars that fail among six cars tested significant, determined by a cutoff value of
0)05?
A) Yes
B) No

Question

The following table describes the results of roadworthiness tests of Ford Focus cars that are three years old (based on data from the Department of Transportation). The random variable x represents the number of cars
That failed among six that were tested for roadworthiness: $$\begin{array} { | c | c | } 
\hline x & P ( x ) \
\hline 0 & 0.377 \
\hline 1 & 0.399 \
\hline 2 & 0.176 \
\hline 3 & 0.041 \
\hline 4 & 0.005 \
\hline 5 & 0 + \
\hline 6 & 0 + \
\hline
\end{array}$$ Is the probability of getting three or more cars that fail among six cars tested significant, determined by a cutoff value of
0)05?
A) Yes
B) No

Quizplus · Accepted Answer

To determine if the probability of getting three or more cars that fail among six cars tested is significant, we need to calculate the probability of getting three or more failures and compare it to the cutoff value of 0.05.

P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
P(X ≥ 3) = 0.041 + 0.005 + 0 + 0
P(X ≥ 3) = 0.046

Since 0.046 is less than 0.05, we can conclude that the probability of getting three or more cars that fail among six cars tested is not significant at the 0.05 cutoff value. Therefore, the answer is A) Yes.

The Following Table Describes the Results of Roadworthiness Tests of Ford