Among live deliveries the probability of a twin birth is 003
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.) 
 | 
Solution
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]
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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]

