1. 1. Fahad
2. 2. A competitor claims that its mid size car gets better mileage than automakers new mid size model. The automaker used sample information in a probability based on normal distribution to provide strong evidence that the competitors claim is false.
3. 3. Total Population 10000 Sample selected 49 of company Y According to EPA (environmental protection agency) standards The mean = 31.55 Standard Deviation = 0.7
4. 4. Well use the normal probability distribution to prove that the competitors claim is false. Secondary data will be used. Sample of 49 mileages has been collected from company Y cars. Mean of sample mileage =31.5 miles/gallon Standard deviation = 1.5 miles/gallon
5. 5. Competing company = 33 = 0.7 Suppose that the competitor is true in his argument P(32 x 35)
6. 6. As it was supposed that the claim is true we therefore take the values to be 32 and 35 instead of 32.3 and 33.7
7. 7. P(32 x 35) = P P(32 x 35) = P(-1.43 z 2.86) From z table(-1.42=.4236) & (2.86=.4979) P(32 x 35)=P .4236+.4979=.9215x100=92.15% This probability says that if the competing automaker claim is valid then 92.15% of all its midsize cars will get mileages between 32mpg and 35mpg
8. 8. 30.8 30.9 32.0 32.3 32.6 31.7 30.4 31.4 32.7 31.4 30.1 32.5 30.8 31.2 31.8 31.6 30.3 32.8 30.6 31.9 32.1 31.3 32.0 31.7 32.8 33.3 32.1 31.5 31.4 31.5 31.3 32.5 32.4 32.2 31.6 31.0 31.8 31.0 31.5 30.6 32.0 30.4 29.8 31.7 32.2 32.4 30.5 31.1 30.6
9. 9. Adding all cars mileage We get 1546.1 As weve taken 49cars sample so 1546.1/49= 31.55 So Mean = 31.55
10. 10. 28 29 30 31 32 33 34 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 MPG
11. 11. The antipollution department and tax department agreed upon Automakers producing cars with mileage of 31.5mpg will be tax-free This will reduce pollution Will be more economical Customers will be attracted to buy it
12. 12. Suppose that an independent testing agency randomly selects one of these cars and finds that it gets mileage of 31.2 mpg when tested. Claimed mean = 33 Standard Deviation =0.7 Which contradicts with the randomly selected car According to the claimed mean and standard deviation the manufacturer should be declared tax-free. Itll be proved that the manufacturers claim is false
13. 13. To calculate the mileage x of a randomly selected car which will be equal to or less than 31.2mpg p( x 31.2) The area under the normal curve to the left side of 31.2 with mean 33.0 and standard deviation 0.7 z= = = -2.57 Which shows that the mileage 31.2 is 2.57 standard deviation below the mean mileage = 33
14. 14. Calculation summary p(x31.2)=p( ) =p(z -2.57)=.5 - .4949=.0051 =.0051x10000 =51.0 Cars mileage
15. 15. It is very hard to believe that 51 cars out of 10000 will have mileage below 31.2 Conclusion: We have a very strong evidence against the competing automakers claim to be false. Reasons could be: Mean is smaller than 33 Standard deviation is greater than 0.7 The population is not normally distributed