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. find the probability of getting three or more cars that fail among six cars tested. note that 0 + represents a positive probability value less than 0.0005.

x p(x)
0 0.377
1 0.399
2 0.176
3 0.041
4 0.005
5 0+
6 0+

a. 0.222
b. 0.005
c. 0.042
d. 0.046

Explanation:

Step1: Identify required probabilities

We need to find $P(x\geq3)$, which is $P(3)+P(4)+P(5)+P(6)$.

Step2: Substitute values

$P(3) = 0.041$, $P(4)=0.005$, $P(5)=0 +$ (less than 0.0005, we can approximate it as 0), $P(6)=0 +$ (approximate as 0).
$P(x\geq3)=0.041 + 0.005+0+0$.

Step3: Calculate result

$P(x\geq3)=0.046$.

Answer:

D. 0.046