Question

Binomial Events Past records over a long period of time show that 10.0% of the screws manufactured by a particular machine ar

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Answer #1

Result:

n=10, p=0.10

a).

mean = np = 1

b).

Variance = np(1 - p) = 0.9

Standard deviation = 0.9487

c).

P(X=x) = (nCx) px (1-p)n-x

Binomial Probabilities Table

X

P(X)

0

0.3487

1

0.3874

2

0.1937

3

0.0574

4

0.0112

5

0.0015

6

0.0001

7

0.0000

8

0.0000

9

0.0000

10

0.0000

d).

Histogram 0.4500 0.3874 0.4000 0.3487 0.3500 0.3000 0.2500 P(X) 0.1937 0.2000 0.1500 0.1000 0.0574 0.0500 0.0112 -0.00150.000

e).

10% of the screws of 10 = 1

Therefore P( x=1)

=0.3874

f).

P( atleast 2 screws defect) = P( x ≥2) = 1- P( x <2)

=1-(P( x=0)+P( x=1))

=1-(0.3487+0.3874)

=0.2639

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