Download 1. Use the Standard Normal Distribution table to find the indicated

1. Use the Standard Normal Distribution table to find the indicated area under the standard
normal curve. (References: example 3 - 6 pages 244 - 247, end of section exercises 21 – 40 page
249) (1 points per each part)
a. Between z = 0 and z = 2.24-
By table lookup:
z = 0, p = 0.5
z = -2.24, p = 0.0125
0.5 – 0.0125 = 0.4875
b. To the left of z = 1.09By table lookup:
z = -1.09, p = 0.1379
c. Between z = -1.15 and z = -0.56
By table lookup:
z = -1.15, p = 0.1251
z = -0.56, p = 0.2877
Subtracting the probabilities:
0.2877 – 0.1251 = 0.1626
d. To the right of z = -1.93
By table lookup
z = -1.93, p = 0.0268
To get the area to the right, subtract p from 1
1 - 0.0268 = 0.9732
Section
5.2: Normal Distributions: Find Probabilities (References: example 1 and 2 page 253, end of
section exercises 13 - 30 pages 257 - 259 2.
The diameters of a wooden dowel produced by a new machine are normally distributed with a
mean of 0.55 inches and a standard deviation of 0.01 inches. What percent of the dowels will
have a diameter greater than 0.57? (5 points)
z = (x - mu)/sigma
z = (0.57 - 0.55)/0.01
z = (0.02)/0.01
z =2
p = 0.9772
That’s the area to the left, so to get the percentage to the right, we subtract from 1
1 – 0.9772 = 0.0228
3. A loan officer rates applicants for credit. Ratings are normally distributed. The mean is 240
and the standard deviation is 50. Find the probability that an applicant will have a rating greater
than 260. (5 points) Section
z = (x - mu)/sigma
z = (260 - 240)/50
z = (20)/50
z =0.4
p = 0.6554
That’s the area to the left, so to get the percentage to the right, we subtract from 1
1 – 0.6554 = 0.3446
5.3: Normal Distributions: Finding Values 4. Answer the questions about the specified normal
distribution. (References: example 4 and 5 page 264 - 265, end of section exercises 39 – 45
pages 267 - 268)
a. The lifetime of ZZZ batteries are normally distributed with a mean of 265 hours and a standard
deviation ? of 10 hours. Find the number of hours that represent the the 25th percentile. (3
points)
using a standard z-table, p = 0.25 corresponds to a z score of -0.67
x = mu + (z*sigma)
x = 265 + (-0.67*10)
x = 265 + -6.74
x = 258.26
b. Scores on an English placement test are normally distributed with a mean of 36 and standard
deviation ? of 6.5. Find the score that marks the top 10%. (3 points)
The top 10% = 90th percentile
using a standard z-table, p = 0.9 corresponds to a z score of 1.28
x = mu + (z*sigma)
x = 36 + (1.28*6.5)
x = 36 + 8.333
x = 44.333
Section 5.4: Sampling Distribution and the Central Limit Theorem 5. Find the probabilities.
(References: example 5 and 6 page 276 - 277, end of section exercises 25 - 31 pages 280 - 281)
a. From National Weather Service records, the annual snowfall in the TopKick Mountains has a
mean of 92 inches and a standard deviation ? of 12 inches. If the snowfall from 25 randomly
selected years are chosen, what it the probability that the snowfall would be greater than 95
inches? (5 points)
z = (x – mu)/(sigma/sqrt(n))
z = (95 – 92)/(12/sqrt(25))
z = 3/(12/5)
z = 1.25
Using a standard z table, p = 0.8944. That’s the probability on the left side of the curve. To get
“greater than”, subtract from 1.
1 – 0.8944 = 0.1056
b. The loan officer rates applicants for credit. Ratings are normally distributed. The mean is 240
and the standard deviation is 60. If 36 applicants are randomly chosen, what is the probability
that they will have a rating between 220 and 260? (5 points)
Where xbar = 220:
z = (x - mu)/(sigma/√n)
z = (220 - 240)/(60/√36)
z = (-20)/10
z =-2
p = 0.0228
Where xbar = 260:
z = (x - mu)/(sigma/√n)
z = (260 - 240)/(60/√36)
z = (20)/10
z =2
p = 0.9772
To get the area between the two, subtract the probabilities.
0.9772 – 0.0228 = 0.9545