Question

Suppose the American Kennel Club estimates that 36% of households in U.S. have a Pembroke Welsh Corgi. A random sample of 400 households is taken. What is the probability the sample proportion is within .03 of P? Question 14 options: 1) .9044 2) 1 3) 0 4) .7887

Answer 1

Solution

Given that,

p = 0.36

1 - p = 1 - 0.36 = 0.64

n = 400

$\mu$ $\hat p$ = p = 0.36

$\sigma$ $\hat p$   $\sqrt$[p ( 1 - p ) / n] = $\sqrt$ [(0.36 * 0.64) / 400 ] = 0.024

P(0.33 < $\hat p$ < 0.39)

= P[(0.33 - 0.36) / 0.024 < ( $\hat p$ - $\mu$ $\hat p$ ) / $\sigma$ $\hat p$ < (0.39 - 0.36) / 0.024 ]

= P(-1.25 < z < 1.25)

= P(z < 1.25) - P(z < -1.25 )

Using z table,   

= 0.8944 - 0.1057

= 0.7887

correct option is = 4