26. 1.X is a normally normally distributed variable with mean 亮 = 30 and standard
deviation
= 4. Find
a) P(x < 40)
b) P(x > 21)
c) P(30 < x < 35)
2.A radar unit is used to measure speeds of cars on a motorway. The speeds are
normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr.
What is the probability that a car picked at random is travelling at more than 100
km/hr?
3.For a certain type of computers, the length of time bewteen charges of the
battery is normally distributed with a mean of 50 hours and a standard deviation
of 15 hours. John owns one of these computers and wants to know the probability
that the length of time will be between 50 and 70 hours.
4.Entry to a certain University is determined by a national test. The scores on this
test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score
better than at least 70% of the students who took the test. Tom takes the test and
scores 585. Will he be admitted to this university?
27. 5. The length of similar components produced by a company are
approximated by a normal distribution model with a mean of 5 cm and a
standard deviation of 0.02 cm. If a component is chosen at random
a) what is the probability that the length of this component is between
4.98 and 5.02 cm?
b) what is the probability that the length of this component is
between 4.96 and 5.04 cm?
28. Answers to the Above QuestionsNote: What is meant here by area is the
area under the standard normal curve.
a) For x = 40, the z-value z = (40 - 30) / 4 = 2.5
Hence P(x < 40) = P(z < 2.5) = [area to the left of 2.5] = 0.9938
b) For x = 21, z = (21 - 30) / 4 = -2.25
Hence P(x > 21) = P(z > -2.25) = [total area] - [area to the left of -2.25]
= 1 - 0.0122 = 0.9878
c) For x = 30 , z = (30 - 30) / 4 = 0 and for x = 35, z = (35 - 30) / 4 = 1.25
Hence P(30 < x < 35) = P(0 < z < 1.25) = [area to the left of z = 1.25] - [area
to the left of 0]
= 0.8944 - 0.5 = 0.3944
29. 2. Let x be the random variable that represents the
speed of cars. x has 亮 = 90 and = 10. We have to
find the probability that x is higher than 100 or
P(x > 100)
For x = 100 , z = (100 - 90) / 10 = 1
P(x > 90) = P(z >, 1) = [total area] - [area to the
left of z = 1]
= 1 - 0.8413 = 0.1587
The probability that a car selected at a random
has a speed greater than 100 km/hr is equal to
0.1587
30. 3. Let x be the random variable that represents the length
of time. It has a mean of 50 and a standard deviation of
15. We have to find the probability that x is between
50 and 70 or P( 50< x < 70)
For x = 50 , z = (50 - 50) / 15 = 0
For x = 70 , z = (70 - 50) / 15 = 1.33 (rounded to 2
decimal places)
P( 50< x < 70) = P( 0< z < 1.33) = [area to the left of z =
1.33] - [area to the left of z = 0]
= 0.9082 - 0.5 = 0.4082
The probability that John's computer has a length of
time between 50 and 70 hours is equal to 0.4082.
31. 4. Let x be the random variable that represents the scores. x
is normally ditsributed with a mean of 500 and a standard
deviation of 100. The total area under the normal curve
represents the total number of students who took the
test. If we multiply the values of the areas under the curve
by 100, we obtain percentages.
For x = 585 , z = (585 - 500) / 100 = 0.85
The proportion P of students who scored below 585 is
given by
P = [area to the left of z = 0.85] = 0.8023 = 80.23%
Tom scored better than 80.23% of the students who took
the test and he will be admitted to this University.
32. 5. a) P(4.98 < x < 5.02) = P(-1 < z < 1) = 0.6826
b) P(4.96 < x < 5.04) = P(-2 < z < 2) = 0.9544
33. 7. What length of time marks the shortest 70%
of all pregnancies?
Normal Distribution 袖 = 266 = 16
P(X < ?) = 0.70 P(Z < ?) = 0.70
Z = 0.52
X = 266 + 0.52(16)
X = 266 + 8.32
X = 274.32
34. For each question, construct a normal distribution curve
and label the horizontal axis. Then answer each
question.
The mean life of a tire is 30 000 km. The standard
deviation is 2000 km.
1. 68% of all tires will have a life between ______km and
______km.
2. 95% of all tires will have a life between ______km and
______km.
3. What percent of the tires will have a life that exceeds
26000 km?
4. If a company purchased 2000 tires, how many tires
would you expect to last more than 28 000 km?