73% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 42 owned dogs are randomly selected, find the probability that

Question

a. Exactly 31 of them are spayed or neutered. ______

b. At most 33 of them are spayed or neutered. _______

c. At least 28 of them are spayed or neutered. _________

d. Between 28 and 32 (including 28 and 32) of them are spayed or neutered. ____________

Answer 1

mean = np = 0.73 * 42 = 30.66

SD = np(1-p) = 2.88

n(1-p) = 42 * 0.27 = 11.34

x = number spayed

since np and n(1-p) both exceed 10, can use normal approximation to binomial distribution

z = (x - mean)/SD = (x - 30.66)/2.88 is distributed N(0,1)

a.

P(x = 31) = P(30.5 < x < 31.5) = P((30.35-30.66)/2.88 < z < (31.5-30.66)/2.88) = P(-0.056 < z < 0.291) = P(z < 0.291) - P(z < -0.056)

left to you to look up probabilities in standard normal distribution table

b.

P(x <= 33) = P(x < 33.5) = P(z < (33.5 - 31.5)/2.88) = P(z < 0.694)

c.

P(x >= 28) = P(x > 27.5) = P(z > (27.5 - 31.5)/2.88) = P(z > -1.39) = 1 - P(z < -1.39)

d.

P(28 <= x <= 32) = P(27.5 < x < 32.5) = P((27.5-31.5)/2.88 < z < (32.5-31.5)/2.88) = P(-1.39 < z < 0.347) = P(z < 0.347) - P(z < -1.39)