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Unformatted text preview: BES Tutorial Sample Solutions Unit 3
1. Among employed women, 25% have never been married. Select 12 employed women
(a) The number in your sample who have never been married has a binomial
distribution. What are the binomial parameters n and p?
(b) What is the probability that exactly 2 of the 12 women in your sample have
never been married? (c) What is the probability that 2 or fewer have never been married? 2. A telemarketer is employed to telephone 10 households each evening between 6 and
7pm with a view to selling a particular service. From past experience it is known that
the probability of any one household being interested in purchasing the service is p =
0.20. Let the random variable X represent the number of households called on a given
evening that are interested in the service.
(a) Comment on the suitability of modelling the (relative frequency) distribution of
X as a binomial distribution.
Recall that we can summarize the precise requirements for a binomial experiment as
follows: There are n identical and independent trials. There are only two possible outcomes for each trial: success and failure. The probability of a success p is the same for each trial.
Is it reasonable to suppose our example satisfies all these conditions ? Independence is often the most problematic assumption but if we’re drawing from a
large market it seems a reasonable assumption here. Some people get annoyed at telemarketers and hang up before they ask their
questions. This is not necessarily a problem here if we define failure as all outcomes
other than having a stated interest in purchasing. While some households may be more predisposed to the particular servic e e.g. richer
households may have a different p from poorer ones; this couldn’t be determined
beforehand. With random sampling (so that one is likely to ask both rich and poor
households) then the constant p represents an average over the whole population.
(b) Use the binomial distribution tables to calculate: P( X = 4), P(X < 4), P(X ≥ 1). We have n=10, p=.2 Obviously can calculate probabilities from either form of the tables. Here we have
simply used is the more convenient form in each case.
3. A believer in the “random walk” theory of stock markets thinks that an index of stock
prices has a probability of 0.65 of increasing in any one year. Let X be the number of
years among the next 5 years in which the index rises.
(a) What do we need to assume for X to have a binomial distribution. What are n
and p? What are the possible values that X can take? Each of the n=5 years represent identical and indepe ndent trials. For this to be true
the fact that the index went up last year does not change the probability it will go up
in the subsequent year.
Index going up is denoted a success and all other outcomes (going down or staying
unchanged)d are taken to be a failure.
The probability of a success p=0.65 is the same for each trial. (b) x Assuming X has a binomial distribution, construct the probability distribution of
X and draw the associated probability histogram. 0
0.3124 Stock index increases over 5 years
0.00 0 1 2 3 Number of increases (c) What are the mean and standard deviation of X? 4 5 5
0.1160 (d) Let Y be the number of years in which the index falls. Assuming X is binomial is
Y binomial? What are the mean and standard deviation of Y? Because all that is being done is to reclassify the outcomes into success and failures Y will
also be binomial. (Ignore the unlikely complication of the index staying the same.) 4. In each of the following cases X is a count. Does it have a binomial distribution?
Justify you answer in each case.
(a) You observe the sex of the next 20 babies born at a local hospital; X is the
number of girls among them.
Yes Recall our concerns about the gender composition example used in lectures: the
fertility decision about whether to have a third child or not being impacted by the
gender mix of the first two children; possibility of multiple births. These are not
(b) A couple decides to continue to have children until their first girl is born; X is
the total number of children the couple has. No Here n is not fixed and depends on previous trial outcomes. 5. The amount of petrol sold daily by a service station is known to be uniformly
distributed between 4,000 and 6,000 litres. What is the probability of sales on any one
day being between 5,500 and 6,000 litres? 6. A software random number generator is designed to produce numbers within a
specified range. We can consider any number in the range as a possible outcome.
Suppose the random number Y can be any value between 0 and 2. Then the uniform
density of the outcomes of Y will have a constant height between 0 and 2 and be zero
(a) What is the height of the density? Draw a graph of the density. Uniform density for random number
0 0 0.5 1 1.5 2 y (b) Find the following: P(Y<0.8); P(Y≤0.8); P(0.5<Y<1.5). 7. Using your personalized project data what is the mean of the variables difficulty and
time? Does it make sense to use the mean to characterize difficulty? What would be a
better way to summarize this variable?
For the fictitious student with SID 1234567 the mean of time is 33.413. This number will
differ across all students because of the personalized nature of the data.
The mean of difficulty is 1.833. (This is the same for all students)
The variable is ordinal but numerical values arbitrarily assigned and hence a better
summary would be to simply determine the proportion of jobs assigned each of the
categories: 1=easy ; 2=standard ; 3=hard . 8. From several years’ records, a fish market manager has determined that the weight of
deep sea bream sold in the market (X) is approximately normally distributed with a
mean of 420 grams and a standard deviation of 80 grams. Assuming this distribution
will remain unchanged in the future, calculate the expected proportions of deep sea
bream sold over the next year weighing
(a) between 300 and 400 grams. (b) between 300 and 500 grams. (c) more than 600 grams. 9. In a certain large city, household annual incomes are considered approximately
normally distributed with a mean of $40,000 and a standard deviation of $6,000.
What proportion of households in the city have an annual income ove r $30,000? If a
random sample of 60 households were selected, how many of these households would
we expect to have annual incomes between $35,000 and $45,000?
) So 95.25% of households in the city have annual incomes greater than $30,000. Therefore we expect 0.5934(60)≈36 households in the sample to have annual incomes
between $35,000 and $45,000. 10. In a certain city it is estimated that 40% of households have access to the internet. A
company wishing to sell services to internet users randomly chooses 150 households
in the city and sends them advertising material. For the households contacted:
(a) Calculate the probability that less than 60 households have internet access?
Let X be the number of households contacted that have inter net access. Then assume X is a
binomial random variable with n=150 and p=0.4. Because n is large we can use the
normal approximation to the binomial where: Thus incorporating the continuity correction we need to find: (b) Calculate the probability that between 50 and 100 (inclusive) households have
internet access? (c) Calculate the probability that more than 50 households have internet access? (d) There is a probability of 0.9 that the number of households with internet access
equals or exceeds what value? There is a 90% chance that the number of households with internet access is 52 or more.
11. The manufacturer of a particular handmade article takes place in two stages. The time
taken for the first stage is approximately normally distributed with a mean of 30
minutes and a standard deviation of 4 minutes. The time taken for the second stage is
also approximately normally distributed, but with a mean of 10 minutes and a standard
deviation of 3 minutes. The times to complete the two stages of production are
independently distributed. Note that the sum of two normally distributed random
variables is also normally distributed.
Let X1 be the time taken for the first stage and X 2 the time taken for the second stage.
Then Y = X1 + X2 is the time taken to complete an article.
(a) What are the mean and standard deviation of the total time to manufacture the
article? Note that independence is used in the calculation of the Var(Y) (b) What is the probability of finishing an article in less than 35 minutes? (c) What proportion of articles will be completed in 35-45 minutes? 12. What is the 25th percentile of the normal distribution N(10, 9)?
Let x be the required percentile. First find z, the 25 th percentile of a standard normal. ...
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- Three '08