Assume a binomial probability distribution has

p = 0.60

and

n = 300.

(a)

What are the mean and standard deviation? (Round your answers totwo decimal places.)

mean standard deviation

(b)

Is this situation one in which binomial probabilities can beapproximated by the normal probability distribution? Explain.

No, because np ≥ 5 and n(1 − p) ≥ 5.Yes, because n ≥ 30.     Yes, becausenp ≥ 5 and n(1 − p) ≥ 5. No, becausenp < 5 and n(1 − p) < 5. Yes,because np < 5 and n(1 − p) <5.

(c)

What is the probability of 160 to 170 successes? Use the normalapproximation of the binomial distribution to answer this question.(Round your answer to four decimal places.)

(d)

What is the probability of 190 or more successes? Use the normalapproximation of the binomial distribution to answer this question.(Round your answer to four decimal places.)

(e)

What is the advantage of using the normal probabilitydistribution to approximate the binomial probabilities?

The advantage would be that using the normal probabilitydistribution to approximate the binomial probabilities increasesthe number of calculations. The advantage would be that using thenormal probability distribution to approximate the binomialprobabilities makes the calculations lessaccurate.     The advantage would be that usingthe normal probability distribution to approximate the binomialprobabilities makes the calculations more accurate. The advantagewould be that using the the normal probability distribution toapproximate the binomial probabilities reduces the number ofcalculations.

How would you calculate the probability in part (d) using thebinomial distribution. (Use f(x) to denote thebinomial probability function.)

P(x ≥ 190) =f(191) + f(192) +f(193) + f(194) +   + f(300)

P(x ≥ 190) =f(190) + f(191) +f(192) + f(193) +   + f(300)

    

P(x ≥ 190) =f(0) + f(1) +    +f(188) + f(189)

P(x ≥ 190) =f(0) + f(1) +    +f(189) + f(190)

P(x ≥ 190) = 1 −f(189) − f(190) −f(191) − f(192) −   − f(300)