Blood type AB is found in only 3 of the population If 170 pe

Blood type AB is found in only 3% of the population†. If 170 people are chosen at random, find the probability of the following. (Use the normal approximation. Round your answers to four decimal places.)(a) 5 or more will have this blood type. (b) between 5 and 10 will have this blood type

Solution

Getting the mean standard deviation of this normal approximation,

u = n p = 5.1

s = sqrt(n p (1 - p)) = 2.224185

Thus, for 5 or more, the critical value is 4.5:

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    4.5      
u = mean =    5.1      
n = sample size =    1 (we treat the whole group as just 1 sample)      
s = standard deviation =    2.224185      
          
Thus,          
          
z =    -0.269761733      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.269761733   ) =    0.606328218 [ANSWER]

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b.

For between 5 and 10 inclusive, the bounds would be 4.5 to 10.5.

Thus, e first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    4.5      
x2 = upper bound =    10.5      
u = mean =    5.1      
n = sample size =    1      
s = standard deviation =    2.224185      
          
Thus, the two z scores are          
          
z1 = lower z score =    -0.269761733      
z2 = upper z score =    2.427855597      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.393671782      
P(z < z2) =    0.992405805      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.598734023 [ANSWER]