NAME______Statistics

Prof. Miller

Problems on Normal Distribution

Directions: Draw a clearly labeled diagram and then solve each problem.

1. The gestation period (length of pregnancy) for male babies born in New York is normally distributed with a mean of 39.4 weeks and a standard deviation of 2.43 weeks.

(a) What percent of mothers of male babies are pregnant for less than 35 weeks?

Z = (35-39.4)/2.43 = -1.81. P(Z<-1.81)= 3.515%

(b) What percent of mothers of male babies are pregnant for between 35 and 40 weeks?

Z = (40-39.4)/2.43 = 0.25. P(-1.81<Z<0.25) = 0.5987 - 0.03515 = 0.56355 = 56.355%

(For this one, I want you to draw a single graph and shade in what is the white area of these two graphs)

(c)What percent of mothers of male babies are pregnant for more than 40 weeks?

P(Z>0.25) = 1 - 0.5987 =0.4013 = 40.13%

(d) Suppose a sailor had shore leave 46 weeks ago and his wife is delivering a baby today. What should the sailor say to his wife when he gets home?

Z = (46-39.4)/2.43 = 2.72. P(Z>2.72) = 0.0033. He should ask her whose the baby is, as there is less than a 1/3% chance that the baby is his.

  1. Suppose that a school district wants to start a program for gifted students. The participants in the program are to be chosen on the basis of IQ scores (normal with  = 100,  = 15). If the school district wants only the top 2% of students to participate in the program, what should be the IQ score that students must exceed to be accepted?

P((X-100)/15) = 0.98

(X-100)/15 = 2.0537

x-100 = 30.81

x = 130.81

  1. The daily demand for gas at Good’s Gas station is normally distributed with a mean of 1812 gallons and a standard deviation of 254 gallons.

(a) What is the probability that the demand for gas will exceed 2000 gallons on any day?

(2000-1812)/254 = 0.74. P(Z<0.74)=0.77035. 1-0.77035 = 0.22965

(b) What is the probability that the demand for gas in a day will be between 1500 and 2000 gallons?

(1500-1812)/254 = -1.228. P(Z<-1.228) = 0.1097. 0.77035 - 0.1097 = 0.66065

(c) What is the probability that the demand for gas will exceed 1500 gallons on any day?

0.22965 + 0.66065 = 0.8903

d) how much gasoline should the station have on hand at the beginning of the day so that the probability of running out of gas that day is only 1%?

P((x-1812)/254)=0.99

(x-1812)/254 = 2.3263

x-1812 = 593.42

x = 2504.42

4. 10% of Americans are left handed.

(a) If 6 people are selected at random, what is the probability that more than 3 of them are left-handed?

(6! / 4! / 2!)(0.1)4(0.9)2 = 15 * 0.0001 * 0.81 = 0.001215

(6! / 5! / 1!)(0.1)5(0.9)1 = 6 * 0.00001 * 0.9 = 0.000054

(6! / 6! / 0!)(0.1)6(0.9)0 = 1 * 0.000001 * 1 = 0.000001

0.001215 + 0.000054 + 0.000001 = 0.00127

(b) Suppose a group of 600 mathematicians get together for a conference. What is the probability that more than 80 of them are left-handed? (Use the normal approximation to the binomial)

Z = (80 - 600*.1) / √(600*.1*.9) = (80-60)/√54 = 20/7.35 = 2.72. P(z<2.72)= 0.9967. P(Z>2.72)=0.0033

5. Airlines sell more tickets for a flight than the number of available seats (overbooking). They do this because they know from past experience that only 90% of ticketed passengers actually show up for the flight. A plane has 325 seats. If the airline sells 350 tickets for a flight, what is the probability that the flight will be overbooked (the number of passengers who show up is greater than the number of available seats)?

Z = (325 - 350 * .9) / √(350*.9*.1) = (325-315) / √(31.5) = 10/5.61 = 1.78. P(z<1.78)=.9625. P(z>1.78) = 0.0375

6. The mean pulse rate for adults is 72 beats per minute ( and let’s suppose that the standard deviation is 10. Find:

a. The probability that a randomly chosen adult has a pulse rate over 80 assuming that the rates are normally distributed.

(80-72)/10 = 0.8. P(z<0.8)=0.8416. P(Z>0.8)=0.1584

  1. The probability that a random sample of 19 adults will have a mean pulse rate over 80.

Z = (80-72) / (10/√19) = 8 / 2.294 = 3.487. P(Z<3.487)=0.9998, so P(Z>3.487) = 0.0002