Tut sol week7

# Tut sol week7 - BES Tutorial Sample Solutions, S1/10 WEEK 7...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: BES Tutorial Sample Solutions, S1/10 WEEK 7 TUTORIAL EXERCISES (To be discussed in the week starting April 19) 1. 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. 300 300 400 80 1.5 0 0.4332 0.3345 (b) between 300 and 500 grams. 300 300 500 420 (c) more than 600 grams. 600 0.25 500 420 1.5 0.3413 600 420 80 1 0 420 80 2.25 0.5 0 0.5 0.4878 0.0122 420 80 0.25 1.5 0 0.0987 80 1.5 0 0.4332 0.7745 400 2.25 1 2. 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 over \$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? ~ 30000 40000, 6000 ) 30000 40000 6000 1.67 0.5 0 1.67 0.5 0.4525 0.9525 So 95.25% of households in the city have annual incomes greater than \$30,000. 35000 40000 45000 40000 35000 45000 6000 6000 0.83 0.83 20 0.83 2 0.2967 0.5934 Therefore we expect 0.5934(60)≈36 households in the sample to have annual incomes between \$35,000 and \$45,000. 3. 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 internet 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: 150 0.4 60 1 150 0.4 0.6 36 Thus incorporating the continuity correction we need to find: 60 59 59.5 59.5 60 6 0.083 0.5 0 0.083 0.5 0.0319 0.4681 (b) Calculate the probability that between 50 and 100 (inclusive) households have internet access? 50 (c) 100 49.5 100.5 49.5 60 100.5 60 6 6 1.75 6.75 0 1.75 0 6.75 0.4599 0.5 0.9599 Calculate the probability that more than 50 households have internet access? 51 0.5 50.5 50.5 60 6 1.583 0.4429 0.9429 (d) There is a probability of 0.9 that the number of households with internet access equals or exceeds what value? 0.9 0 0.5 0.5 60 0.9 6 60.5 0.4 6 60.5 1.28 52.82 6 0.9 There is a 90% chance that the number of households with internet access is 52 or more. 4. 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 X2 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? 30 5 ~ 3 4 40,25 10 40 25 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? 35 35 40 5 1 0.5 0 0.5 0.3413 0.1587 1 (c) What proportion of articles will be completed in 35-45 minutes? 35 45 35 1 40 5 2 0 0.6826 45 1 40 5 1 5. What is the 25th percentile of the normal distribution N(10, 9)? Let x be the required percentile. First find z, the 25th percentile of a standard normal. 0.25 0.25 0 0.5 .25 0.25 0.67 0.67 ~ 10,9 25 : 10 0.67 3 7.99 ...
View Full Document

## This note was uploaded on 09/22/2011 for the course ECON 1203 taught by Professor Denzilgfiebig during the Three '11 term at University of New South Wales.

Ask a homework question - tutors are online