The circumference of a circle is 15.2mm.
2 answers:
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Answer:
0.0307
Step-by-step explanation:
Given data;
Rivets from vendor A = 70%; = 0.7
Rivets from vendor B = 80%; = 0.8
Number of rivets purchased from each vendor (n) =500
Assuming that and
represents the mean numbers of rivets that hits the targeted specification.
Using the Central Limit Theorem to evaluate the Binomial distribution of and
; Then:
= 500 × 0.7
= 350
> 10
= 500 × 0.3
= 150
> 10
= 500 × 0.8
= 400
> 10
= 500 × 0.2
= 100
> 10
Hence;
≅ N (
,
) = N (350,105)
≅ N (
,
) = N (400, 80)
T = +
T ≅ N (750, 185)
We are tasked to find out that what is the probability that more than 775 of the rivets meet the specifications; So, we need to determine that P(T > 775); the continuity correction point T = 775.5
The corresponding Z-value is as follows:
Z= 1.87
P(T>775) =P(Z>1.87)
= 1 - 0.9693
= 0.0307
∴ the probability that more than 775 of the rivets meet the specifications = 0.0307