Among live deliveries, the probability of a twin birth is 0.03. In 2,478 live de
ID: 3378695 • Letter: A
Question
Among live deliveries, the probability of a twin birth is 0.03.
In 2,478 live deliveries, what is the probability of at least 84 twin births? (Use Excel or table for calculation of probability. Round standard deviation to 2 decimal places. Round your answer to 4 decimal places.)
Fewer than 69? (Use Excel or table for calculation of probability. Round standard deviation to 2 decimal places. Round your answer to 4 decimal places.)
(a)In 2,478 live deliveries, what is the probability of at least 84 twin births? (Use Excel or table for calculation of probability. Round standard deviation to 2 decimal places. Round your answer to 4 decimal places.)
)Fewer than 69? (Use Excel or table for calculation of probability. Round standard deviation to 2 decimal places. Round your answer to 4 decimal places.)
Explanation / Answer
a)
Here,
mean = n p = 2478*0.03 = 74.34
standard deviation = sqrt[n p(1-p)] = sqrt(2478*0.03*(1-0.03)) = 8.49
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 83.5
u = mean = 74.34
s = standard deviation = 8.49
Thus,
z = (x - u) / s = 1.078916372
Thus, using a table/technology, the right tailed area of this is
P(z > 1.078916372 ) = 0.140312505 [ANSWER]
**********************
b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 68.5
u = mean = 74.34
s = standard deviation = 8.49
Thus,
z = (x - u) / s = -0.68786808
Thus, using a table/technology, the left tailed area of this is
P(z < -0.68786808 ) = 0.24576793 [ANSWER]
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