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Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked.

probability x is at most 30? less than 30? between 15 and 25?

Respuesta :

taskmasters
p = 0.11
1-p = .89
n = 200

mean = n*p = 22
std dev = n*sqrt(p(1-p)/n) = 4.425

find z values for standard normal distribution
Z= (30-22)/4.425 = 1/808
look up table to find P(Z less than 1.808) which is same P(x less than 30)

for 2nd part, you need 2 Z values
Z1 = (25-22) / 4.425 = 0.678
Z2 = (15-22) / 4.425 = -1.58

P(-1.58 less than Z less than 0.678) = P(Z less than .678) - P(Z less than - 1.58)

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