Question

Anyone who bets money in a casino is classified as a winner if he or she wins more than he or she loses. Casino operators in Atlantic City, New Jersey, believe that the proportion of all players who go home a winner is 0.46. Suppose 75 Atlantic City gamblers are selected at random. a. [3 marks] Find the sampling distribution of the proportion of gamblers who go home winners. b. [2 marks] Find the probability that the sample proportion is less than 0.40. c. [2 marks] Find the probability that the sample proportion is more than 0.45. d. [3 marks] Find a symmetric interval about the mean (p=0.46) such that the probability that the sample proportion is in this interval is 0.99

Answer 1

Solution

Given that,

p = 0.46

1 - p = 1 - 0.46 = 0.54

n = 75

a)   = p = 0.46

=  [p ( 1 - p ) / n] =   [(0.46 * 0.54) / 75 ] = 0.0575

b) P( < 0.40)

= P[( - ) / < (0.40 - 0.46) / 0.0575 ]

= P(z < -1.04)

Using z table,

= 0.1492

c) P( > 0.45) = 1 - P( < 0.45 )

= 1 - P(( - ) / < (0.45 - 0.46) / 0.0575 )

= 1 - P(z < -0.17)

Using z table

= 1 - 0.4325

= 0.5675

d) Using standard normal table,

P( -z < Z < z) = 0.99

= P(Z < z) - P(Z <-z ) = 0.99

= 2P(Z < z) - 1 = 0.99

= 2P(Z < z) = 1 + 0.99

= P(Z < z) = 1.99 / 2

= P(Z < z) = 0.995

= P(Z < 2.58) = 0.995

= z  ± 2.58

Using z-score formula,

  = z *   +  

  = -2.58 * 0.0575 + 0.46

  = 0.31

Using z-score formula,

  = z *   +  

  = 2.58 * 0.0575 + 0.46

  = 0.61

The probability is 0.99 that the sample percentage will be contained above 0.31 and below 0.61