A dart is to be thrown at a target. The probability the dart will hit the target (yes or no) on a single attempt is 0.20. Each throw is independent of the other throws. Let X be the number of attempts before the target is hit.
a. What type of distribution does X follow?
b. Compute the expected number of attempts needed before hitting the target?
c. What is the variance and standard deviation of the number of attempts needed before hitting the target?
d. What is the probability that it will take less than 5 throws before hitting the target?

a) X has geometric distribution with parameter 0.2

b) The expected number of attempts before hitting the target is 5

c) The variance is 20, the standard deviation is √20

d) The probability that less than 5 throws are needed before hitting the target is 0.5904.

Step-by-step explanation:

a) X has geometric distribution, with parameter p=0.2.

b) If X is geometric with parameter p, then its mean is 1/p and the variance is . With p = 0.2, the mean is 1/0.2 = 5. As a consecuence, the expected number of attempts before hitting the target is 5.

c) The variance with p = 0.2 is 0.8/0.2² = 20. The standard deviation is √20.

d) The probability that it takes less than 5 throws before hitting the target can be computed with the probability of the complementary event: the probability that the first 4 targets miss. That probability is 0.8⁴=0.4096. Therefore, the probability that it will take less than 5 throws before hitting the target is 1-0.4096 = 0.5904.

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !