Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

This will help you in your revision as it has questions and answers

Typology: Exercises

2021/2022

1 / 40

Download MANCOSA Business statistics practice Questions and more Exercises Business Statistics in PDF only on Docsity! Mancosa Business statistics
practice Questions and Answers
sig
HONORIS UNITED UNIVERSITIES
Class Exercize 1 1. The number of bicycles sold monthly by a bicycle dealer was: 25, 18, 30, 18, 20, 19, 30, 16, 36, 24 a) Find the mean and median number of bicycles sold monthly. b) Find the range, variance and standard deviation of the number of bicycles sold monthly. 2. The number of houses sold in an upmarket area of Cape Town in 2012 along with the price range (in Millions of Rands is shown in the table below: Price (Millions of Rands) Houses sold 1.2 < 1.4 4 1.4 < 1.6 9 1.6 < 1.8 18 1.8 < 2.0 26 2.0 < 2.2 17 2.2.< 2.4 6 Determine the: 2.1. Mean price 2.2. Median price 2.3. Modal price 2.4. Standard Deviation ....... ....... Business Statistics BBA/B.Com Solution for No.1a ....... • Formula: ∑x / n = 25+18+30+18+20+19+30+16+36+ 24 / 10 • = 236 / 10 • = 23,6 • Mode= 18 and 30 • Median= 22 • Tip: for median arrange the data first in ascending order ....... ....... Business Statistics BBA/B.Com Solution for No.1b ....... • Range= 36-16 = 20 • 𝑆2 = ∑𝑥2 ̶ 𝑛(𝑥 )2 𝑛 ̶ 1 • = 5962 ̶ 10(23.60)2 10−1 • 5962 ̶ 5569,6 9 • =43,61 • S= 6.603 ....... ....... Business Statistics BBA/B.Com Problem no.2 ....... The number of houses sold in an upmarket area of Cape Town in 2012 along with the price range (in Millions of Rands is shown in the table below: . Price (Millions of Rands) Houses sold 1.2 < 1.4 4 1.4 < 1.6 9 1.6 < 1.8 18 1.8 < 2.0 26 2.0 < 2.2 17 2.2 < 2.4 6 ....... ....... Business Statistics BBA/B.Com Solution for No.2 ....... • 𝑆𝐷 = ∑𝑓𝑚2 ̶ 𝑛(𝑥 )2 𝑛 ̶ 1 = 279.6 ̶ 80(1.85)2 79 = 0.271 million Exercise 2: Probabilities 1. Global Insurance has found that 1 in 5 of all insurance policies are surrendered (cashed in) before the maturity date. Assume that 10 policies are randomly selected from the company’s policy database. What is the probability that 4 of these 10 insurance policies will be surrendered before their maturity date? 2. From the same problem, what is the probability at least two out of the 10 randomly selected policies will be surrendered before their maturity date? 3. An outdoor furniture store distributes flyers at the 2016 Homemakers Expo. There is a 70% chance that a customer will require to have his purchase delivered to his home. If eight customers are randomly surveyed, what is the probability that fewer than three customers will take the item home with them (do not require to have the item delivered)? 4. A family consists of 3 children. What is the probability that at most 2 of the children are boys? 5. Peter owns a hardware store and he decided to boost his sales by distributing flyers at a very busy crossing in Pretoria. There is a 65% chance that a customer will be over the age of 50. If 8 customers are randomly surveyed, what is the probability that more than 5 customers will be over 50 years old? Exercise 2 Solutions Exercise 2 Solutions 4. At most 2 = P(0)+P(1)+P(2) p= 0.5 q= 0.5 n=3 P(0) = 3C0 x 0.50 x 0.53 = 0.125 P(1) = 3C1 x 0.51 x 0.52 = 0.375 P(2) = 3C2 x 0.52 x 0.51 = 0.375 Therefore at most 2= 0.125+0.375+0.375= 0.875…..hence a 87.5% chance Exercise 2 Solutions 5. P(r˃5) = P(6) + P(7) + P(8) n=8 p= 0.65 q= 0.35 P(6)= 8C6 x 0.656 x 0.352 = 0.259 P(7)= 8C7 x 0.657 x 0.351 = 0.137 P(8)= 8C8 x 0.658 x 0.350 = 0.032 Therefore P(r˃5) = 0.259+0.137+0.032= 0.428…..hence a likelihood of 42.8% Class Exercise 3 1. A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes. 2. Find the probability that the shopper will be in the store more than 39 minutes 3. A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed. 4. A coffee dispensing machine used in cafetarias is set to dispense coffee with an average fill of 230 ml and a standard deviation of 10 ml per cup. Assume that the volume dispensed is normally distributed. For a randomly selected cup dispensed by the machine, what is the probability that: a) The cup is filled to more than 235 ml? b) The cup is filled to between 235 ml and 245 ml 5. Assume that the purchase value of transactions, x, at a national clothing store such as Edgars is normally distributed with a mean of R244 and a standard deviation of R68. a) What is the minimum purchase value of transactions for the highest spending 15% of clothing store customers? b) What purchase value of transactions separates the lowest spending 20% of clothing store customers from the remaining customers? Exercize 3 Solutions 3. 𝑃 𝑥 < 2 = 𝑃 𝑧 < 2 −2.4 0.5 = 𝑃 𝑧 < −0.8 = 0.5 − 0.2881 = 𝟎. 𝟐𝟏𝟏𝟗 4a) 𝑃 𝑥˃235 = 𝑃 𝑧 < 235 −230 10 = 𝑃 𝑧˃0.5 = 0.5 − 0.1915 = 𝟎. 𝟑𝟎𝟗 b) 𝑃 235 < 𝑥 < 245 = 𝑃( 235 −230 10 < 𝑧 < 245 −230 10 ) = 𝑃 0.5 < 𝑧 < 1.5 = 0.4332 − 0.1915 = 𝟎. 2417 Exercise 3 Solutions 5. Z=(X-)/ • X= ? Z = ? = R244 = R68 • Z = upper 15% will be from 35 % upwards since standard normal distribution graph is symmetrical with two halves equaling 50% each. Z score for 35% = 1.04 • Therefore if we rearrange our formulae to make X the subject we get: X = + Z Exercise 3 Solutions • X = R244+(1.04 × R68) = R244 + R70.72 = R314.72 • Thus the highest spending 15% of clothing store customers spend at least R314.72 per transaction b) Z = lower 20%. Lower 20% lies in the negative side of the distribution. To get z score for lower 20% we need to find z score for 30% for negative half of the distribution. When consulting the table of z scores we find that z score = -0.84 Y
MANCOSA
B.Com ITM
Module : Advanced Business Statistics
....... ....... Advanced Business Statistics B.Com ITM Index Numbers Problem ....... 1. Telkom offers a range of telecommunication services to small businesses. A small printing business has used the services of TalkPlus (a value added telephone service), SmartAcess (an advertising service) and ISDN (an internet connection) for the past three years. Their annual usage and unit price for each service are given in the following table: ....... ....... Advanced Business Statistics B.Com ITM Problem 1 Data ....... Telkom Services 2009 2010 2011 Unit price P0 Quantity(10 0s calls) Q0 Unit Price Quantity(100s calls) Unit Price Quantity(100s calls) TalkPlus 65 14 70 18 55 17 SmartAcess 35 27 40 29 45 24 ISDN 50 16 45 22 40 32 ....... ....... Advanced Business Statistics B.Com ITM 2009 vs 2010 (Laspeyres) ....... p0 x q0 p1 x q0 910 980 945 1080 800 720 2655 2780 ....... ....... Advanced Business Statistics B.Com ITM 2009 vs 2011 (Laspeyres) ....... p0 x q0 p1 x q0 910 770 945 1215 800 640 2655 2625 ....... ....... Advanced Business Statistics B.Com ITM Composite Price Index Solution ....... . ....... ....... Advanced Business Statistics B.Com ITM Composite Quantity Index ....... • c) Composite Quantity Index (Paasche): (2010): 3410/2780 x 100 = 122.67 (2011): 3295/2625 x 100 = 125.52 • d) Overall usage of different telecommunication services has increased. Regression and Correlation Exercise An electronics store in Cape Town, has recorded the number of tablets sold each week and the number of advertisements placed weekly for a period of 12 weeks. Advertisement Tablets Sold 90 96 85 98 75 85 69 65 62 60 49 65 55 59 50 50 30 45 35 50 42 48 50 55 Calculate the linear regression equation Using the calculated linear regression, predict how many tablets will be sold if 60 advertisements were placed. Calculate the Pearson Correlation co-efficient BACHELOR OF COMMERCE IN INFORMATION AND TECHNOLOGY MANAGEMENT Module : Advanced Business Statistics