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UNIT 6 HOMEWORK SOLUTIONS
Math 102 & Core 143
- Complete the following table regarding tossing a fair coin and counting the number of heads.
| Number of heads | Percent of heads Number of tosses | Expected value | SEsum | Expected value | SE% 100 | 50 | 5 | 50% | 5% 2500 | 1250 | 25 | 50% | 1% 10000 | 5000 | 50 | 50% | .5% 1000000 | 500000 | 500 | 50% | .05%
- A box contains 1 zero and 9 ones. True or false and justify your answer.
i) If we draw 10 times we have a better chance of getting exactly 1 zero than the chance of getting
exactly 10 zeros on 100 draws.
True. It is easier to get something exactly when we have fewer trials.
ii) If we draw 100 times we have a better chance of getting exactly 10 zeros than the chance of
getting exactly 100 zeros on 1000 draws.
True. It is easier to get something exactly when we have fewer trials.
iii) If we draw 10 times we will get closer to 10% zeros than if we draw 10000 times.
False. It is easier to get close to the expected percentage with more trials. This is the law of averages.
- In gambling, the following is a commonly held belief. If a gambler has lost 10 times in a row,
then the gambler should continue playing because he is due to win, by the law of averages. Explain
why this reasoning is incorrect.
The law of averages tells us nothing about “the next draw.” Here, each game is independent of the
previous game. Consider flipping a fair coin 10 times. If all flips are tails, is the chance that the
next flip is a head more than 50%? No. It’s still 50%.
- Create box models for each of the following situations.
i) We roll 20 dice and want to add the sum of the value rolled.
The box has the numbers 1,2,3,4,5,6 and we draw 20 times.
ii) We roll a pair dice and win $1 if the sum is greater than or equal to 10. We lost $1 otherwise.
This game is played 50 times.
Box has 6 1’s and 30 − 1 ’s as there are 6 ways (out of 36) to have a sum of 10 or more.
- True or false: We flip a fair coin 100 times. The chance that we get exactly 50 heads is less than
the chance that we get exactly 50% heads.
False, they are the same thing.
- Consider a box with the numbers 1,2,7,9,9,10 in it. We draw 100 times (with replacement).
i) What is the chance that the sum of draws is less than 100?
0%. Since the minimum sum is 100, we can’t get less than 100.
ii) What is the chance that the sum of draws is greater than 1000?
0%. Since the maximum sum is 1000, we can’t get more than 1000.
iii) What is the approximate chance that the sum of draws is between 650 and 750?
Here we need EV and SE. We find EV ≈ 633. 33 and SE ≈
√ 1+4+49+81+81+ 6
− (6.33)^2 ≈
- Using standard units, the area under the Normal curve from. 47 to 3. 29 is ≈ 31%.
We bet $1 on roulette and have a 12 38 chance^ of^ winning.^ If^ we^ win^ we^ win^ $2^ (otherwise,^ we^ lose our $1). We play this game 100 times. Fill in the blanks:
We expect to win/lose (circle one) $ , give or take $.
lose $5.26; $13.94; box has 12 2 ’s and 26 − 1 ’s; first blank is EV and the second is SE.
8. You roll a die 180 times and count the number of !•^ ’s.
i) Fill in the blanks: We expect to get!
’s., give or take!
30; 5. The box has one 1 and 5 0 ’s.
ii) If we gather a large group of people to each do the above experiment, about what percentage of
these people should get an answer between 15 and 45?
99 .73%; convert 15 and 45 to standard units (-3 and 3) and use the Normal curve.
- A box has the numbers − 1 , 0 , 1 in it. We make 400 draws from this box. Approximate the
chance that our sum is
This box has average and SD =
√ (−1)^2 +0^2 +1^2 3 −^0
(^2) ≈. 8165. Hence, we know the mean and standard
deviation of the distribution we are sampling from.
From this we have SEsum =
a) greater than 0
We want the area under the Normal curve to the right of. 5 which is
. 5 − 0 16. 3 ≈^.^03 in^ standard^ units. Looking this up on the Normal curve, we have about 48 .80%.
b) at least 10
We want the area under the Normal curve to the right of 9. 5 which is
- 5 − 0
- 3 ≈^.^58 in^ standard^ units. This gives an answer of ≈ 28 .10%.
c) less than − 15
We want the area under the Normal curve to the left of − 15. 5 which is − 15. 5 − 0
3 ≈^ −.^95 ,^ giving answer of ≈ 17 .11%.
A coin is twice as likely to land heads as it is tails.
a) Use the Central Limit Theorem to approximate the chance that on 90 flips I will get exactly 60
heads.
We have μ = 90 · 23 = 60 and SE =
√ 90(2/3)(1/3) ≈ 4. 47. Using the Normal curve we want the
area under it from 59.5 to 60.5. Converting these to standard units gives about −. 11 and. 11. From
the Normal chart, the answer is ≈ 8 .76%.
b) Use the Binomial formula to find the exact chance of the event given in part (a).
The exact answer is
( 90 60
) ( 2 3
) 60 ( 1 2
) 30 which gives ≈ 8 .89%.
- The average person weights 150 pounds with an SD of 35 pounds. We want to lift 50 people in
