Question

2. If 20% of the memory chips made in a certain plant are defective, what are

the probabilities that in a lot of 100 randomly chosen for inspection

(a)At most 15 will be defective

(b)Exactly 15 will be defective

(c)More than 18 will be defective

Answer 1

​​​​​​ANSWER:

a)

X ~ Bin ( n , p)

Where n = 100 , p = 0.20

Using Normal Approximation to Binomial
Mean = n * P = ( 100 * 0.2 ) = 20
Variance = n * P * Q = ( 100 * 0.2 * 0.8 ) = 16
Standard deviation = √(variance) = √(16) = 4

P(X < x) = P(Z < ( x - mean ) / SD )

P ( X <= 15 ) = ?
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 15 + 0.5 ) = P ( X < 15.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 15.5 - 20 ) / 4
Z = -1.13
P ( ( X - µ ) / σ ) < ( 15.5 - 20 ) / 4 )
P ( X < 15.5 ) = P ( Z < -1.13 )
P ( X < 15.5 ) = 0.1292

b)

P ( X = 15 ) = ?
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 15 - 0.5 < X < 15 + 0.5 )

= P ( 14.5 < X < 15.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 14.5 - 20 ) / 4
Z = -1.38
Z = ( 15.5 - 20 ) / 4
Z = -1.13
P ( -1.38 < Z < -1.13 )
P ( 14.5 < X < 15.5 ) = P ( Z < -1.13 ) - P ( Z < -1.38 )
P ( 14.5 < X < 15.5 ) = 0.1292 - 0.0838
P ( 14.5 < X < 15.5 ) = 0.0454

c)

P ( X > 18 ) = ?
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 18 + 0.5 )

= P ( X > 18.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 18.5 - 20 ) / 4
Z = -0.38
P ( ( X - µ ) / σ ) > ( 18.5 - 20 ) / 4 )
P ( Z > -0.38 )
P ( X > 18.5 ) = 1 - P ( Z < -0.38 )
P ( X > 18.5 ) = 1 - 0.3520
P ( X > 18.5 ) = 0.6480

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