Question

1.A sample of 80 is drawn from a population with a proportion equal to 0.50. Determine the probability of observing between 33 and 50 successes.

2.For a population that is left skewed with a mean of 27 and a standard deviation equal to 16​, determine the probability of observing a sample mean of 25 or more from a sample of size 37.

3. For a normal population with a mean equal to 80 and a standard deviation equal to 11​, determine the probability of observing a sample mean of 87 or less from a sample of size 11.

Answer 1

1)

Solution:

Given that,

P = 0.50

1 - P = 0.50

n = 80

Here, BIN ( n , P ) that is , BIN (80 , 0.50)

then,

n*p = 80 * 0.50 = 40 > 5

n(1- P) = 80 * 0.50 = 40 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 40

Standard deviation = =n*p*(1-p) = 80 * 0.50 *0.50 = 20

We using continuity correction factor

P(32.5 < x < 50.5) = P((32.5 - 40)/ 20) < (x - ) /  < (50.5 - 40) / 20) )

= P(-1.68 < z < 2.35)

= P(z < 2.35) - P(z < -1.68)

= 0.9906 - 0.0465

Probability = 0.9441

2)

Solution :

Given that ,

mean = = 27

standard deviation = = 16

n = 37

= = 27 and

= / n = 16 / 37 = 2.6304

P( 25) = 1 - P( 25)

= 1 - P(( - ) / (25 - 27) / 2.6304)

= 1 - P(z -0.76)

= 1 - 0.2236 Using standard normal table.

Probability = 0.7764

3)

Given that ,

mean = = 80

standard deviation = = 11

n = 11

= = 80 and

= / n = 11 / 11 = 3.3166

P( 87) = P(( - ) / (87 - 80) / 3.3166)

= P(z 2.11)

= 0.9826 Using standard normal table,    

Probability = 0.9826