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 .04 of P?

Question 14 options:

	1) 1
	2) .9044
	3) 0
	4) .7887

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Answer 1

Solution

Given that,

p = 0.36

1 - p = 1 - 0.36 = 0.64

n = 400

= p = 0.36

  [p ( 1 - p ) / n] = [(0.36 * 0.64) / 400] = 0.024

P(0.32 < < 0.40 )

= P[(0.32 - 0.36) / 0.024 < ( - ) / < (0.40 - 0.36) / 0.024]

= P(- 1.667 < z < 1.667)

= P(z < 1.667 ) - P(z < - 1.667 )

Using z table,   

= 0.9522 - 0.0478

= 0.9044

correct option is = 2)