Math 141 Extra Practice over New material for Final.
Note: The final Exam is comprehensive, so you have to review all the
topics listed in "Concepts to Know"
1. Let Z be the standard normal random variable. Find a if:
(a) P (Z < a) = 0.234
(b) P (−a < Z < a) = 0.286
2. Firestarters Inc. manufactures artificial starter logs for
fireplaces. These logs are accepted by the buyer if they fall within the
tolerance limits of .695 inches and .780 inches in length. Assuming that
the length of the logs is normally distributed with a mean of 0.72
inches and a standard deviation of 0.03 inches, estimate the percentage
of logs that will be rejected by the buyer.
3. The weight of a length of rope will support is normally
distributed with a mean of 2000 lbs and a standard deviation of 50 lbs.
What is the probability that you will buy a rope that can support
between 1900 lbs and 2050 lbs?
4. An engineering class takes an exam in which the average was
65.7 and the standard deviation 12.93. The prof decides to give F’s to
8% of the class, D’s to 18% of the class, B’s to 25% of the class, A’s to
10% of the class, and C’s to the rest. What is the lowest grade you can
make and still pass (assuming C or higher is passing).
5. The average GPA for the graduating class at Bryan High
School this year was a 1.96 with a standard deviation of 0.68. What is
the lowest GPA a student in the graduating class could have and still
be in the top 25% of the class?
6. Let Z be the standard normal random variable. Find
P(Z<1.35).
7. Let Z be the standard normal random variable. Find a if
(a) P (Z > a) = 0.432
(b) P (Z < a) = 0.268
8. A tire manufacturer claims that the life of its tires, calculated
in miles, is a normally distributed random variable with a mean of
24,000 and a standard deviation of 1,400 miles. What is the probability
that a tire will last more than 25,000 miles?
9. The random variable X is normally distributed with a mean of
65 and a standard deviation of 6. Find the percent of area under the
normal curve that is within 1.5 standard deviations of the mean.
10. The life span of a 60 watt light bulb is normally distributed
with an average life span of 8,000 hours and a standard deviation of 15
days. What is the probability that a bulb selected at random will last
at least 8,250 hours?
11. A bank deposit paying simple interest at the rate of 4.5%/year
grew to a sum of $1800 in 10 months.
Find the principal.
12. Suppose you invest $11,000 into an account paying 7.2%
interest compounded weekly. How many years will it take until you
have $15,000 in the account?
13. Mark recently put $2400 down on a new truck and financed
the rest of the amount at 5.8% interest compounded monthly over the
next 5 years. If his monthly payments are set at $350/month,
(a) What was the original cash price of the truck?
(b) How much interest will Mark end up paying on the truck?
14. What is the effective yield on an account that pays nominal
interest rate of 3.947% compounded daily?
15. Suppose you invest $2,000 into an account paying 6.35%
interest compounded quarterly. How much
money would you have after 8 years?
16. A bank deposit paying simple interest at the rate of 5%/year
grew to a sum of $3100 in 10 months. Find the principal.
17. Find the accumulated amount at the end of 9 months on a
$1000 bank deposit paying simple interest at a rate of 8%/year.
18. How much money should William invest in an account
compounded daily at a rate of 7.5%, if he wants to have $20,000 in 30
years?
19. Suppose you invest $12,000 into an account paying 7.6%
interest compounded weekly. How many years will it take until you
have $15,000 in the account?
20. The Hill’s recently bought a house for $145,000. They put
$5000 down and financed the rest at 6.2% interest compounded
monthly for the next 30 years.
(a) What will there monthly payments be?
(b) How much interest will they end up paying on their house?
(c) If they decided to pay $100 more toward the principle of the
loan each month, how long will it take them to pay off their house?
(d) How much money will they end up saving (in the end) if they
choose to pay $100 more each month?
Answers
1. (a) invnorm(0.234,0,1)=-0.7257; (b) invnorm(0.643,0,1)=0.3665
2. 0.2251 obtained by: P(rejected) = 1-P(accepted)=
1-normalcdf(.695,.780,.72,.03)
3. 0.8186
4. 57.3815
5. 2.4187
6. 0.9115
7. (a) 0.1713, (b) -0.6189
8. 0.2375
9. 0.8664
10. 0.2437
11. $1734.94
12. 4.31 years
13. (a) $20,591.32
14. 4.026%
15. $3,310.69
16. $2976
17. $1060
18. $2108.47
19. t=2.98 years
20. (a) $857.46 (b) $168,685.60 (c) t=22.78 years (d) $46,954.33