Question

Question 1:

A binomial distribution has p = 0.21 and n = 94. Use the normal approximation to the binomial distribution to answer parts (a) through (d) below. Round to four decimal places as needed.)

a. What are the mean and standard deviation for this distribution?

b. What is the probability of exactly 15 successes?

c. What is the probability of 13 to 21 successes?

d. What is the probability of 9 to 17 successes?

Question 2:

The average number of miles driven on a full tank of gas in a certain model car before its low-fuel light comes on is 331. Assume this mileage follows the normal distribution with a standard deviation of 44 miles. Complete parts a through d below. (Round to four decimal places as needed.)

1. What is the probability that, before the low-fuel light comes on, the car will travel less than 361 miles on the next tank of gas?
2. What is the probability that, before the low-fuel light comes on, the car will travel more than 280 miles on the next tank of gas?
3. What is the probability that, before the low-fuel light comes on, the car will travel between 301 and 321 miles on the next tank of gas?
4. What is the probability that, before the low-fuel light comes on, the car will travel exactly 353 miles on the next tank of gas?

Answer 1

1) n = 0.21

    p = 94

a) mean() = np = 94 * 0.21 = 19.74

Standard deviation() = sqrt(np(1 - p))

                               = sqrt(94 * 0.21 * (1 - 0.21))

                              = 3.9490

b) P(X = 15)

= P(14.5 < X < 15.5)

= P((14.5 - )/< (X - )/< (15.5 - )/)

= P((14.5 - 19.74)/3.9490 < Z < (15.5 - 19.74)/3.9490)

= P(-1.33 < Z < -1.07)

= P(Z < -1.07) - P(Z < -1.33)

= 0.1423 - 0.0918

= 0.0505

c) P(13 < X < 21)

= P(12.5 < X < 21.5)

= P((12.5 - )/< (X - )/< (21.5 - )/)

= P((12.5 - 19.74)/3.9490 < Z < (21.5 - 19.74)/3.9490)

= P(-1.83 < Z < 0.45)

= P(Z < 0.45) - P(Z < -1.83)

= 0.6736 - 0.0336

= 0.6400

d) P(9 < X < 17)

= P(8.5 < X < 17.5)

= P((8.5 - )/< (X - )/< (`17.5 - )/)

= P((8.5 - 19.74)/3.9490 < Z < (17.5 - 19.74)/3.9490)

= P(-2.85 < Z < -0.57)

= P(Z < -0.57) - P(Z < -2.85)

= 0.2843 - 0.0022

= 0.2821

2)a) P(X < 361)

= P((X - )/ < (361 - )/)

= P(Z < (361 - 331)/44)

= P(Z < 0.68)

= 0.7517

b) P(X > 280)

= P((X - )/ > (280 - )/)

= P(Z > (361 - 331)/44)

= P(Z > -1.16)

= 1 - P(Z < -1.16)

= 1 - 0.1230

= 0.8770

c) P(301 < X < 321)

= P((301 - )/ < (X - )/ < (321 - )/)

= P((301 - 331)/44 < Z < (321 - 331)/44)

= P(-0.68 < Z < -0.23)

= P(Z < -0.23) - P(Z < -0.68)

= 0.4090 - 0.2483

= 0.1607

d) P(X = 353) = 0