Step-by-step answer:
Given:
mean, mu = 200 m
standard deviation, sigma = 30 m
sample size, N = 5
Maximum deviation for no damage, D = 100 m
Solution:
Z-score for maximum deviation
= (D-mu)/sigma
= (100-200)/30
= -10/3
From normal distribution tables, the probability of right tail with
Z= - 10/3
is 0.9995709, which represents the probability that the parachute will open at 100m or more.
Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.
P(all five safe) = 0.9995709^5 = 0.9978565
So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m
Answer:
the base is 10 and the height is 6.4
The probability of event A and B to both occur is denoted as P(A ∩ B) = P(A) P(B|A). It is the probability that Event A occurs times the probability that Event B occurs, given that Event A has occurred.
So, to find the probability that you will be assigned a poem by Shakespeare and by Tennyson, let Event A = the event that a Shakespeare poem will be assigned to you; and let Event B = the event that the second poem that will be assigned to you will be by Tennyson.
At first, there are a total of 13 poems that would be randomly assigned in your class. There are 4 poems by Shakespeare, thus P(A) is 4/13.
After the first selection, there would be 13 poems left. Therefore, P(B|A) = 2/12
Based on the rule of multiplication,
P(A ∩ B) = P(A) P(B|A)P(A ∩ B) = 4/13 * 2/12
P(A ∩ B) = 8/156
P(A ∩ B) = 2/39
The probability that you will be assigned a poem by Shakespeare, then a poem by Tennyson is 2/39 or 5.13%.
