Question

Using the normal distribution, calculate the following probabilities:

1. a) P(X≤16|n=50, p=0.70)

2. b) P(10≤X≤16|n=50, p=0.50)

Answer 1

Solution:

Given that,

a)

P = 0.70

1 - P = 0.30

n = 50

Here, BIN ( n , P ) that is , BIN (50 , 0.70)

then,

n*p = 50*0.70 = 35 > 5

n(1- P) = 50*0.30 = 15 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 35

Standard deviation = =n*p*(1-p) = 50*0.70*0.30 = 10.5

We using countinuity correction factor

P( X a ) = P(X < a + 0.5)

P(x < 16.5) = P((x - ) / < (16.5 - 35) / 10.5)

= P(z < -5.70 )

= 0

Probability = 0

b)

P = 0.50

1 - P = 0.50

n = 50

Here, BIN ( n , P ) that is , BIN (50 , 0.50)

then,

n*p = 50*0.50 = 25 >5

n(1- P) = 50*0.50 = 25 >5

According to normal approximation binomial,

X Normal

Mean = = n*P = 25

Standard deviation = =n*p*(1-p) = 50*0.50*0.50 = 12.5

We using countinuity correction factor

P(10 ≤ X ≤ 16) = P(9.5 < X < 16.5)

= P((9.5 - 25) /12.5 < ((x - ) / < (16.5 - 25) /12.5)

= P(-4.38 < Z < -2.40)

= P(Z < -2.40) - P(Z < -4.38)

= 0.0082 - 0

= 0.0082

Probability = 0.0082