On the Titanic 1514 of the 2224 passengers died Suppose that

On the Titanic, 1514 of the 2224 passengers died.

Suppose that 30% of those onboard came onboard at Cherbourg. What is the probability (correct to three decimal places) that in a random sample of 30 people onboard between 30% and 38% came onboard at Cherbourg? Please show all work.

Solution

Note that

u = mean = n p = 9
s = standard deviation = sqrt(n p (1 - p)) = 2.50998008

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    9      
x2 = upper bound =    11.4      
u = mean =    9      
n = sample size =    30      
s = standard deviation =    2.50998008      
          
Thus, the two z scores are          
          
z1 = lower z score =    0      
z2 = upper z score =    5.237229366      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.5      
P(z < z2) =    0.999999918      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.5000 [ANSWER]