Title: Self test: Probability
Course: UGQuants (SSPS) > Section: Probability

Exercise 1

This self-test has 16 exercises.

1. Using standard normal tables find the z scores corresponding to the following proportion of cases in a normally distributed variable:

a. The 0.683 of cases closest to the mean.

b. The 0.05 of cases with the highest values.

c. The 0.242 of cases with the lowest values.

d. The 0.05 of cases that form the two extremes (low and high) of the distribution.

Exercise 1 - Answer

a. The 0.683 of cases closest to the mean.

z = -1.0 to z = +1.0

b. The 0.05 of cases with the highest values.

z = +1.645 and greater

c. The 0.242 of cases with the lowest values.

z = -0.7 and below

d. The 0.05 of cases that form the two extremes (low and high) of the distribution.

z < -1.96 and z > +1.96

Exercise 2

2. Find the proportion of observations in a normally distributed variable that lie between the following z-scores:

a. Between 0.0 and +2.0

b. Above +1.3

c. Between 0.0 and +1.3

d. Between -0.5 and -3.4

e. Less than -1.96 and greater than +1.96

Exercise 2 - Answer

a. Between 0.0 and +2.0

0.477

b. Above +1.3

0.097

c. Between 0.0 and +1.3

.403

d. Between -0.5 and -3.4

0.3082

e. Less than -1.96 and greater than +1.96

.050

Exercise 3

3. Calculate the probability that a case in a normally distributed variable falls between the following z-scores:

a. Between 0.0 and 1.0

b. Between 0.0 and +1.96

c. Less than -1.96 and greater than 1.96

Exercise 3 - Answer

a. Between 0.0 and 1.0

0.341

b. Between 0.0 and +1.96

0.475

c. Less than -1.96 and greater than 1.96

0.050

Exercise 4

4. The breaking distance of motorists travelling at 50 miles an hour on the open road is found to be normally distributed with a mean of 40 meters and standard deviation of 5 meters.

a. If a pedestrian, unaware of an oncoming car travelling at 50 miles an hour, steps into the road 47 meters from the car, what is the chance that they will be knocked down?

b. And what if they are 30 meters away?

c. How far away would the pedestrian have to be from the car to have a 50/50 chance of not being hit?

Exercise 4 - Answer

The breaking distance of motorists travelling at 50 miles an hour on the open road is found to be normally distributed with a mean of 40 meters and standard deviation of 5 meters.

a. If a pedestrian, unaware of an oncoming car travelling at 50 miles an hour, steps into the road 47 meters from the car, what is the chance that they will be knocked down?

z = (47-40) / 5 = 7/5 = +1.4
If stopping distances are normally distributed, in .5 of all events, the motorist will stop in under 40 meters (i.e. below the mean of 40 meters). If Z is +1.4, in .419 of events they will stop between 40 meters and 47 meters. In the remaining .081 of events they will stop in over 47 meters. The probability will be 0.081, or an 8.1% chance of being knocked down.

b. And what if they are 30 meters away?

z = (30-40) / 5 = -10/5 = -2.0
If stopping distances are normally distributed, in .5 of all events, the motorist will stop in over 40 meters (i.e. above the mean of 40 meters). If Z is +2.0, in .477 of events they will stop between 30 meters and 40 meters. In the remaining .023 of events they will stop in under 30 meters. The probability will be 0.977, or a 97.7% chance of being knocked down.

c. How far away would the pedestrian have to be from the car to have a 50/50 chance of not being hit?

We don't even have to calculate the z score here. They will have an even chance of being hit if they are the mean stopping distance away: half the cars will stop in time, the other half will hit the pedestrian.

Exercise 5

5. The heights of a group of 1000 people are found to be approximately normally distributed with a mean of 1.75 meters and a standard deviation of 10 centimetres. How many people in the group are more than 1.95 meters tall?

Exercise 5 - Answer

The heights of a group of 1000 people are found to be approximately normally distributed with a mean of 1.75 meters and a standard deviation of 10 centimetres. How many people in the group are more than 1.95 meters tall?

z = (195 -175)/ 10 = +2 Therefore .023 of people will be taller than 1.95 meters, or 23 people in one thousand

Exercise 6

6. In a sample of 100 cases, the mean age of respondents is found to be 40 with a standard deviation of 10. What will be the 95% confidence interval for this result?

Exercise 6 - Answer

In a sample of 100 cases, the mean age of respondents is found to be 40 with a standard deviation of 10. What will be the 95% confidence interval for this result?

standard error of the mean = population standard deviation / square root of the sample size
= 10/10
= 1
95% of cases will lie within +/-1.96 standard errors of the mean
Confidence interval will be 38.04 to 41.96

Exercise 7

7. In a sample of 10,000 men, the mean height of respondents is found to be 173.0 centimetres with a standard deviation of 15 centimetres.

a. What will be the 95% confidence interval for this result?

b. For the same sample what will be the 99% confidence interval?

c. Suppose our results came from a sample of only 100 men. What will be the 99% confidence interval?

Exercise 7 - Answer

In a sample of 10,000 men, the mean height of respondents is found to be 173.0 centimetres with a standard deviation of 15 centimetres.

a. What will be the 95% confidence interval for this result?

= 15/100
= .15 centimetres
95% of estimates will lie within +/-1.96 standard errors of the mean = 0.294 cm
Confidence interval will be 172.706 cm to 173.294 cm

b. For the same sample what will be the 99% confidence interval?

99% of estimates will lie within +/-2.58 standard errors of the mean = 0.387 cm
Confidence interval will be 172.613 cm to 173.387 cm

c. Suppose our results came from a sample of only 100 men. What will be the 99% confidence interval?

Standard error of the mean = 15/10 = 1.5
99% of estimates will lie within +/-2.58 standard errors of the mean = 3.87 cm
Confidence interval will be 169.13 cm to 176.87 cm

Exercise 8

8. In a random sample of 2429 adults, 935 are regular smokers. What is the 95% confidence interval for the proportion of smokers in the adult population?

Exercise 8 - Answer

In a random sample of 2429 adults, 935 are regular smokers. What is the 95% confidence interval for the proportion of smokers in the adult population?

Proportion (p) who are smokers: 935 / 2429 = 0.385
SE (p) = √(p*(1-p)) / √n
= √(0.385* (1-0.385)) / √2429
= √0.237 / √2429 = 0.487 / 49.28 = 0.00988
For a confidence interval of 95% we need to include cases within 1.96 standard errors.
So we can be 95% certain that the proportion of smokers in the population will be:
0.385 +/- (1.96 * 0.00988)
= 0.385 +/- 0.019
= 0.366 to 0.404
Or, we could say that between about 37% and 40% of the adult population are likely to be smokers.

Exercise 9

9. If an event has a probability of occurring of 0.4. What is the probability of it not occurring?

- 0.6
Feedback: Correct!
- We cannot know
Feedback: Incorrect
- 0.25
Feedback: Incorrect
- 1.5
Feedback: Incorrect

Exercise 10

10. Which of the following could constitute a 'trial' in the language of probability?

- All of the above are correct
Feedback: Correct!
- Tossing a coin
Feedback: Incorrect
- Drawing a card from a deck
Feedback: Incorrect
- Answering a question in a survey
Feedback: Incorrect

Exercise 11

11. If p=0.1 what does this mean?

- All of the above are true
Feedback: Correct!
- Something has a one in ten chance of happening
Feedback: Incorrect
- Ten out of a hundred trials were successful
Feedback: Incorrect
- The proportion of successful trials in all trials was 0.1
Feedback: Incorrect

Exercise 12

12. A standard normal curve is?

- A normal curve with a standard deviation of 1 and mean of zero
Feedback: Correct!
- A normal curve with a standard deviation of zero and mean of 1
Feedback: Incorrect
- The only normal curve where the standard deviation equals the mean
Feedback: Incorrect
- The only normal curve which is symmetrical
Feedback: Incorrect

Exercise 13

13. A random sample of 10,000 workers gives an estimate for their mean monthly income of 400 dollars. The standard deviation in the sample for monthly income is 50 dollars. What is the 95% confidence interval for the mean, rounded to the nearest whole dollar?

- 399 to 401 dollars
Feedback: Correct!
- 398 to 402 dollars
Feedback: Incorrect
- 390 to 410 dollars
Feedback: Incorrect
- 400 dollars exactly
Feedback: Incorrect

Exercise 14

14. What would the answer be if our sample size was only 100 workers, but the sample mean monthly income (400 dollars) and sample standard deviation (50 dollars) remained the same?

- 390 to 410 dollars
Feedback: Correct!
- 397 to 403 dollars
Feedback: Incorrect
- 398 to 402 dollars
Feedback: Incorrect
- 400 dollars exactly
Feedback: Incorrect

Exercise 15

15. And what if the sample size was 10,000 and the mean monthly income 400 dollars but the standard deviation for the sample was 150 dollars?

- 397 to 403 dollars
Feedback: Correct!
- 390 to 410 dollars
Feedback: Incorrect
- 399 to 401 dollars
Feedback: Incorrect
- 400 dollars exactly
Feedback: Incorrect

Exercise 16

16. Imagine we carry out a random sample survey of the ages of the population in Edinburgh at the same time as the ten year census is carried out (a census is a survey that covers everyone and not just a sample). Our survey estimates the mean age of the population to be 44.5 years, with a 95% confidence interval of ± 1.2 years. When the census results are published however, we discover that the mean age of the population was 42.9 years. What has happened?

- All of the above are true
Feedback: Correct!
- Our estimate happened to lie further than two standard errors from the mean of estimates that would have been obtained from infinitely repeated random samples
Feedback: Incorrect
- The true population value lay outside our confidence interval
Feedback: Incorrect
- We had a 5% or less chance of getting the sample result we obtained
Feedback: Incorrect