Jills bowling scores are approximately normally distributed

Jill’s bowling scores are approximately normally
distributed with mean 170 and standard deviation
20, while Jack’s scores are approximately normally
distributed with mean 160 and standard deviation
15. If Jack and Jill each bowl one game, then
assuming that their scores are independent random
variables, approximate the probability that
(a) Jack’s score is higher;
(b) the total of their scores is above 350.

Please solve using integrals

Solution

a)

We let Jill\'s scores to be population 1, and Jacks\' scores be population 2.

The difference of their scores have a distribution of

u = u1 - u2 = 170-160 = 10

s(x1-x2) = sqrt(S1^2 + s2^2) = 25

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    10      
          
s = standard deviation =    25      
          
Thus,          
          
z = (x - u) / s =    -0.4      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.4   ) =    0.344578258 [ANSWER]

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b)

The mean total of their scores is 170 + 160 = 330.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    350      
u = mean =    330      
          
s = standard deviation =    25      
          
Thus,          
          
z = (x - u) / s =    0.8      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.8   ) =    0.211855399 [ANSWER]