Download 8. Assume the speed of vehicles along a stretch of I

8. Assume the speed of vehicles along a stretch of I-10 has an approximately normal
distribution with a mean of 71 mph and a standard deviation of 8 mph.
a. The current speed limit is 65 mph. What is the proportion of vehicles less than or
equal
to the speed limit?
b. What proportion of the vehicles would be going less than 50 mph?
c. A new speed limit will be initiated such that approximately 10% of vehicles will be over
the speed limit. What is the new speed limit based on this criterion?
d. In what way do you think the actual distribution of speeds differs from a normal
distribution?
Ans.
a. = 71 mph;  = 8 mph
The proportion of vehicles less than or equal to the speed limit 65 mph = 0.2266
b. The proportion of the vehicles would be going less than 50 mph = 0.0043
c. The new speed limit would be 81.25 mph (approximately 81 mph).
d. The actual distribution of speeds on a highway differs from the normal distribution in
that
i. It is not symmetric. The speeds are not symmetrically distributed about the central
vaue. Two speeds that are an equal amount greater or lower than the mean speed, may
not have the same frequency as is required for a normal distribution.
ii. Secondly, the speeds further away from the average speed do not drop in frequency
as rapidly as in normal distribution. The frequency of the extreme values may not be as
low as in normal distribution, resulting in a fatter tail. The distribution may also be
skewed in one direction. This is specially true in highways where the distribution will be
expected to be positively skewed- greater frequency in the higher speed range than
lower.
11. A group of students at a school takes a history test. The distribution is normal with a
mean of 25, and a standard deviation of 4. (a) Everyone who scores in the top 30% of
the distribution gets a certificate. What is the lowest score someone can get and still
earn a certificate? (b) The top 5% of the scores get to compete in a statewide history
contest. What is the lowest score someone can get and still go onto compete with the
rest of the state?
Ans. For the test, the mean score = 25;  = 4
Using the normal calculator (Ch-7, Lane),
a. the lowest score someone can get and still earn a certificate = 27.096 (or approx..
27.1)
b. the lowest score someone can get and still go onto compete with the rest of the state
= 31.58 (or approx. 31.6)
12. Use the normal distribution to approximate the binomial distribution and find the
probability of getting 15 to 18 heads out of 25 flips. Compare this to what you get when
you calculate the probability using the binomial distribution. Write your answers out to
four decimal places.
Ans For the binomial distribution of coin flipping, probability of getting a head
(success) in one flip = p = 0.5
N = no. of trials = 25
Mean is  = N*p = 25 * 0.5 = 12.5
Variance σ2 = Np(1-p) = (25)(0.5)(0.5) = 6.25
The standard deviation is therefore σ = √6.25 = 2.5.
Using the binomial calculator (Lane CH-5),
the probability of getting 15 to 18 heads out of 25 flips = 0.2049
Using the normal distribution to approximate the binomial distribution: The area
between 14.5 and 18.5 in the normal distribution curve is an approximation of the
probability of obtaining 15 to 18 heads.
Area below 14.5 = 0.7881
Area below 18.5 = 0.9918
Therefore,
Area between 14.5 and 18.5 = 0.9918 – 0.7881 = 0.2037
So in this case,
the probability of getting 15 to 18 heads out of 25 flips = 0.2037
Illowsky CH-6
60. The patient recovery time from a particular surgical procedure is normally distributed
with a mean of 5.3 days and a standard deviation of 2.1 days.
What is the median recovery time?
a. 2.7
b. 5.3
c. 7.4
d. 2.1
Ans. b. 5.3
In the normal distribution, mean and median are the same.
66. Height and weight are two measurements used to track a child’s development. The
World Health Organization measures child development by comparing the weights of
children who are the same height and the same gender. In 2009, weights for all 80 cm
girls in the reference population had a mean μ = 10.2 kg and standard deviation σ = 0.8
kg. Weights are normally distributed. X ~ N(10.2, 0.8). Calculate the z-scores that
correspond to the following weights and interpret them.
a. 11 kg
b. 7.9 kg
c. 12.2 kg
Ans. a. z-score for 11 kg = (11- μ)/ σ = (11- 10.2)/ 0.8 = 0.8/0.8 = 1
Interpretation: The weight 11 kg is 1 standard deviation above the average weight value
of 10.2 kg.
b. z-score for 7.9 kg = (7.9 - μ)/ σ = (7.9 - 10.2)/ 0.8 = -2.3/0.8 = -2.875
Interpretation: The weight 7.9 kg is 2.875 standard deviations below the average weight
value of 10.2 kg.
c. z-score for 12.2 kg = (12.2- μ)/ σ = (12.2 - 10.2)/ 0.8 = 2.0/0.8 = 2.5
Interpretation: The weight 12.2 kg is 2.5 standard deviations above the average weight
value of 10.2 kg. As the z-score for this weight is greater than 2, it can be concluded that
12.2 kg weight for a girl 8 cm tall is an “unusual” value.
76. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally
distributed with a mean of 250 feet and a standard deviation of 50 feet.
a. If X = distance in feet for a fly ball, then X ~ _____(_____,_____)
b. If one fly ball is randomly chosen from this distribution, what is the probability that this
ball traveled fewer than 220 feet? Sketch the graph. Scale the horizontal axis X. Shade
the region corresponding to the probability. Find the probability.
c. Find the 80th percentile of the distribution of fly balls. Sketch the graph, and write the
probability statement.
Ans.a. If X = distance in feet for a fly ball, then X ~ _N____(__250___,__50___)
b. z-score for 220 ft = (220- μ)/ σ = (220 - 250)/ 50 = -30/50 = -0.6
Area to the left of z= -0.6 will represent the probability that this ball traveled fewer than
220 feet.
Area to the left of -0.6 = 0.5 – 0.22575 = 0.27425
So the probability that this ball traveled fewer than 220 feet is 0.27425.
c. The 80th percentile of the distribution of fly balls = 292.073 feet
80% of the fly balls travelled fewer than 292.073 feet. 20% of the fly balls travelled
greater than 292.073 feet.
88. Facebook provides a variety of statistics on its Web site that detail the growth and
popularity of the site.
On average, 28 percent of 18 to 34 year olds check their Facebook profiles before
getting out of bed in the morning. Suppose this percentage follows a normal distribution
with a standard deviation of five percent.
a. Find the probability that the percent of 18 to 34-year-olds who check Facebook before
getting out of bed in the morning is at least 30.
b. Find the 95th percentile, and express it in a sentence.
Ans. If X= percentage of 18 to 34 year old who check their Facebook profiles before
getting out of bed in the morning, then X~ N(28, 5).
= 28;  = 5
a. z-score for 30% = (30- )/ = (30-28)/5 = 2/5 = 0.4
The area to the right of z-score 0.4 will represent probability that the percent of 18 to 34year-olds who check Facebook before getting out of bed in the morning is at least 30.
From the normal table (Illowsky pg 849),
Area to the right of 0.4 = 1- (0.5 + 0.15542) = 0.3446
The required probability is 0.3446.
c. 95% of the area will be below that percentage.
Area to the left = 0.95
From the normal table we need to find the area value closest to 0.95 - 0.5 = 0.45 and get
the corresponding z-score. The closest match lower than 0.45 that we can get on the
table is 0.44950 and the corresponding z-score value is 1.64.
Now we convert the z-score 1.64 to the x value using the formula
z*=x-
x = z *  +  = 1.64 *5 + 28 = 36.2
So the 95th percentile is 36.2%.
It means the probability that 36.2 percent of the 18 to 34-year-olds who check Facebook
before getting out of bed in the morning, is 0.95 or less.
CH-7 Illowsky
62. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally
distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly
sample 49 fly balls.
a. If
X ̄ = average distance in feet for 49 fly balls, then
X̄
~ __()
b. What is the probability that the 49 balls traveled an average of less than 240 feet?
Sketch the graph. Scale the horizontal axis for X ̄ . Shade the region corresponding to
the probability. Find the probability.
c. Find the 80th percentile of the distribution of the average of 49 fly balls.
Ans. a. X ̄ ~ _N_(250, 50/√49)
Or X ̄~ N(250, 7.1429)
b. z-score for 240 = (240-250)/7.1429 = -10/7.1429 =-1.4
Probability that the 49 balls traveled an average of less than 240 feet = area left of z
score -1.4 = 0.5 – 0.41924 = 0.0808
c. 80th percentile of the distribution of the average of 49 fly balls = 256.01 (using
normal calculator, Lane)
Calculations using the z score table, Illowsky:
Area 0.80 (= 0.5 + 0.3) corresponds to z-score 0.84
X ̄ = z* x + x = 0.84 * 7.1429 + 250 = 256.00
70. Which of the following is NOT TRUE about the distribution for averages?
a. The mean, median, and mode are equal.
b. The area under the curve is one.
c. The curve never touches the x-axis.
d. The curve is skewed to the right.
Ans. d. The curve is skewed to the right.
96. A typical adult has an average IQ score of 105 with a standard deviation of 20. If 20
randomly selected adults are given an IQ test, what is the probability that the sample
mean scores will be between 85 and 125 points?
Ans.
x =  = 105
x = /√n = 20/ √20 = 4.4721
If X ̄ = average IQ score for 20 adults,
X ̄ ~ N(105, 4.4721)
z-score for 85 = (85-105)/4.4721 =-20/4.4721 = -4.4721
Area o the left of z score -4.4721 = 0
z-score for 125 = (125-105)/4.4721 = 20/4.4721 = 4.4721
Area to the left of z-score 4.4721 = 1
Probability that the sample mean scores will be between 85 and 125 points
= Area between z scores -4.4721 (mean score 85) and +4.4721 (mean score 125)
= Area to the left of z-score 4.4721 - Area to the left of z-score -4.4721
= 1- 0
=1