Question

A random variable Y whose distribution is binomial with parameters are n = 500 and p= 0.400, and here Y suggests the number of desired outcomes of the random experiment and n-Y is the number of undesired outcomes obtained from a random experiment of n independent trials. On this random experiment   p̂ sample proportion is found as Y/n. (Round your answers to 3 decimal places in all parts.)

a)What is the expected value of this statistic?

b)Between what limits will 99% of the sample proportion occur?

c)What is the probability that sample proportion exceed 0.456?

d). What is the probability that sample proportion less than 0.344?

Answer 1

Solution

n = 500

a ) = 0.400

1 - = 1 - 0.400 = 0.600

b ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 * (((0.400 * 0.600) / 500 )

= 0.056

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.400 - 0.056 < p < 0.400 + 0.056

0.456 < p < 0.247

=   = 0.400

= ( 1 - ) / n

= (((0.400 * 0.600) / 500 )

= 0.0219

= 0.0219

C ) P(   > 0.456 )

= 1 - P (   < 0.456 )

= 1 - P ( -    / ) < ( 0.456 - 0.400 / 0.0219 )

= 1 - P ( z < 0.056 / 0.0219 )

= 1 - P ( z < 2.56)

Using z table

= 1 - 0.9948

=0.0052

Probability = 0.0052

D) P (   < 0.344)

P ( -    / ) < ( 0.344 - 0.400 / 0.0219 )

P ( z < - 0.056 / 0.0219 )

Using z table

P ( z < -2.56)

=0.0052

Probability = 0.0052