BES Tutorial Sample Solutions, S2 2010

It will be posted on BES website with one week delay

WEEK 7 TUTORIAL EXERCISES (To be discussed in the week starting August 30)

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 400300 420

80400 420

80

1.5 0.25 0 1.5 0 0.25

0.4332 0.0987 0.3345

(b) between 300 and 500 grams.

300 500300 420

80500 420

80

1.5 1 0 1.5 0 1

0.4332 0.3413 0.7745

(c) more than 600 grams.

600600 420

80

2.25 0.5 0 2.25 0.5 0.4878 0.0122

 

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?

~ 40000, 6000 )

3000030000 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 4500035000 40000

600045000 40000

6000

0.83 0.83 2 0 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?

50 100 49.5 100.5  49.5 60

6100.5 60

6 

1.75 6.75  0 1.75 0 6.75  

0.4599 0.5 0.9599  

(c) Calculate the probability that more than 50 households have internet access?

51 50.5  

50.5 606

 

1.583  0.5 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 0.5 0.9

0.5 60

60.9 

060.5

60.4 

60.56

1.28 52.82 

 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 10 40 

 4 3 25 

5 ~ 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?

3535 40

5

1 0.5 0 1 0.5 0.3413 0.1587

 (c) What proportion of articles will be completed in 35-45 minutes?

35 4535 40

545 40

5

1 1 2 0 1 0.6826

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 : 

103

0.67 

7.99