two percent of hair dryers produced in a certain plant are defective... estimate the probability that of 10,000 randomly selected hair dryers, exactly 225 are defective?
A) 0.0057
B) 0.0051
C) 0.0065
D) 0.0034
D) 0.0367
A) 0.0057
B) 0.0051
C) 0.0065
D) 0.0034
D) 0.0367
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i doubt whether you can easily find a binomial calculator that can handle this.
moreover, as #s are very large, & both np & nq are well above 10, the normal approximation will give sufficiently accurate results.
a. µ = np = 10,000*.02
b. σ = sqrt(npq) = sqrt(10,000*.02*.98)
c. with continuity correction, exactly 225 becomes 224.5 to 225.5
d. z1 = (224.5-µ)/σ , z2 = (225.5-µ)/σ
e. P(x = 225) = P(z1 ≤ z ≤ z2) = 0.0057 [A] <------
moreover, as #s are very large, & both np & nq are well above 10, the normal approximation will give sufficiently accurate results.
a. µ = np = 10,000*.02
b. σ = sqrt(npq) = sqrt(10,000*.02*.98)
c. with continuity correction, exactly 225 becomes 224.5 to 225.5
d. z1 = (224.5-µ)/σ , z2 = (225.5-µ)/σ
e. P(x = 225) = P(z1 ≤ z ≤ z2) = 0.0057 [A] <------
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10000C225 (.02)^225 (.98)^9775
= you calculate
= you calculate