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Answer:
probability that both passes a defective item is 0.8742
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Step-by-step explanation:</h3>
probability that the first inspector misses is Pr( 1st misses)= 0.06
therefore the probability he does not miss is
Pr(1st passes)= 1 - Pr( 1st misses) = 1 - 0.06 = 0.94
probability that the second misses is Pr( 2nd misses) = 0.07
therefore probability that 2nd does not miss is
Pr( 2nd passes) = 1- Pr( 2nd misses) = 0.93
probability that both passes a defective item is Pr(1st passes)*Pr( 2nd passes)
= 0.93*0.94 = 0.8742
Answer:
The minimum sample size needed for use of the normal approximation is 50.
Step-by-step explanation:
Suitability of the normal distribution:
In a binomial distribution with parameters n and p, the normal approximation is suitable is:
np >= 5
n(1-p) >= 5
In this question, we have that:
p = 0.9
Since p > 0.5, it means that np > n(1-p). So we have that:





The minimum sample size needed for use of the normal approximation is 50.