Problem

A consumer organization that evaluates new automobiles customarily reports the number of...

A consumer organization that evaluates new automobiles customarily reports the number of major defects in each car examined. Let X denote the number of major defects in a randomly selected car of a certain type. The cdf of X is as follows

Calculate the following probabilities directly from the cdf:

a. p(2), that is , P(X=2)

b.P(X > 3)

c.P (2 ≤ X ≤ 5)

d. P(2< X < 5)

Step-by-Step Solution

Solution 1

Step #1 of 5

Given below is the cumulative probability distribution of number of major defects in a randomly selected car of a certain type.

The cumulative distribution function is,

$$ F(x)= \begin{cases}0 & \text { if } x<0 \\ 0.06 & \text { if } 0 \leq x<1 \\ 0.19 & \text { if } 1 \leq x<2 \\ 0.39 & \text { if } 2 \leq x<3 \\ 0.67 & \text { if } 3 \leq x<4 \\ 0.92 & \text { if } 4 \leq x<5 \\ 0.97 & \text { if } 5 \leq x<6 \\ 1 & \text { if } x \geq 6\end{cases} $$

Step #2 of 5

(a)

$$ \begin{aligned} p(2) &=P(X=2) \\ &=F(2)-F(1) \\ &=0.39-0.19 \\ &=0.2 \end{aligned} $$

$P(X=2)=0.2$

Step #3 of 5

(b)

$$ \begin{aligned} P(X>3) &=1-P(X \leq 3) \\ &=1-F(3) \\ &=1-0.67 \\ &=0.33 \end{aligned} $$

Step #4 of 5

(c)

$$ \begin{aligned} P(2 \leq X \leq 5) &=F(5)-F(2)+P(X=2) \quad(\text { Since }, P(a< X \leq b)=F(b)-F(a)) \\ &=0.97-0.39+0.20 \\ &=0.78 \end{aligned} $$

Therefore, $P(2 \leq X \leq 5)=0.78$.

Step #5 of 5

(d)

$$ \begin{aligned} P(2< X< 5) &=\{F(5)-F(2)\}-P(X=5) \\ &=\{F(5)-F(2)\}-\{F(5)-F(4)\} \\ &=\{0.97-0.39\}-\{0.97-0.92\} \\ &=0.53 \end{aligned} $$

Therefore, $P(2< X< 5)=0.53$.

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