Below is a probability distribution for the number of failures in an elementary statistics course. X 0 1 2 3 4 P(X=x) 0.41 0.18

Answer:

a) P(X = 2) = 0.29

b) P(X < 2) = 0.59

c) P(X ≤ 2) = 0.88

d) P(X > 2) = 0.12

e) P(X = 1 or X = 4) = 0.24

f) P(1 ≤ X ≤ 4) = 0.59

Step-by-step explanation:

Here’s the probability distribution we have:

P(X = 0) = 0.41

P(X = 1) = 0.18

P(X = 2) = p

P(X = 3) = 0.06

P(X = 4) = 0.06

a. P(X = 2) =

The total probability must sum to 1. Therefore

0.41 + 0.18 + p + 0.06 + 0.06 = 1

which leads us to find that p = 1 - 0.71

Thus, p = 0.29

P(X = 2) = 0.29

b. P(X < 2) =

This value can be computed as P(X < 2) = P(X = 0) + P(X = 1) = 0.41 + 0.18 = 0.59

c. P(X ≤ 2) =

P(X ≤ 2) totals up P(X = 0), P(X = 1), and P(X = 2): 0.41 + 0.18 + 0.29 = 0.88

d. P(X > 2) =

P(X > 2) sums P(X = 3) and P(X = 4): 0.06 + 0.06 results in 0.12.

e. P(X = 1 or X = 4) =

This is calculated as P(X = 1 or X = 4) = P(X = 1) + P(X = 4) = 0.18 + 0.06 = 0.24

f. P(1 ≤ X ≤ 4) =

For this, we calculate P(1 ≤ X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4), equating to 0.18 + 0.29 + 0.06 + 0.06, which sums up to 0.59

Answer:

a) P(X=2)= 0.29

b) P(X<2)= 0.59

c) P(X≤2)= 0.88

d) P(X>2)= 0.12

e) P(X=1 or X=4)= 0.24

f) P(1≤X≤4)= 0.59

Step-by-step explanation:

a) To find P(X=2), we calculate: P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4) which equals 1 - 0.41 - 0.18 - 0.06 - 0.06, resulting in 0.29

b) For P(X<2), we sum P(X=0) and P(X=1): 0.41 + 0.18 yields 0.59

c) To obtain P(X≤2), we add P(X=0), P(X=1), and P(X=2): 0.41 + 0.18 + 0.29 equals 0.88

d) To calculate P(X>2), we find P(X=3) + P(X=4): 0.06 + 0.06 gives us 0.12

e) For P(X=1 or X=4), we use the union of probabilities: P(X=1) + P(X=4) which is 0.18 + 0.06, resulting in 0.24

f) P(1≤X≤4) is found by adding P(X=1), P(X=2), P(X=3), and P(X=4): 0.18 + 0.29 + 0.06 + 0.06 results in 0.59