Eastwood's BA201 Assignments


Homework Answers for Chapter 6 Exercises 21-30

21.       a.         46.41%, found by (20.27 – 20.00)/0.15 = 1.8

            b.         3.59%, found by 0.5000 – 0.4641

            c.         81.85%, found by 0.3413 + 0.4772

            d.         27.43%, found by 0.5000 – 0.2257

 

22.        a.         About 34.13%, found by (34 – 32)/2 = 1.00, the area for 1.00 is 0.3413

            b.         About 4.95%, found by (28.7 – 32)2 = – 1.65.  The area for – 1.65 is 0.4505, then 0.5000 – 0.4505 = 0.0495

            c.         About 77.45%, found by (29 – 32)/2 = – 1.5.  The area for – 1.5 is 0.4332.  Then 0.4332 + 0.3413(from part a) = 0.7745

            d.         About 35.3 hours, found by 1.65 = (X – 32)/2

 

23.        a.         – 0.4 for net sales, found by (170 – 180)/25 and 2.92 for employees, found by (1850 – 1500)/120

            b.         Net sales are 0.4 standard deviations below the mean.  Employees is 2.92 standard deviations above the mean. Draw a normal curve and locate the approximate position of these values.

            c.         65.54% of the aluminum fabricators have greater net sales compared with Clarion, found by 0.1554 + 0.5000.  Only 0.18% have more employees than Clarion, found by 0.5000 – 0.4982

 

24.        a.         15.87%, found by (15 – 20)/5 = – 1.0.  The area for –1.0 is 0.3413.  Then 0.5000 – 0.3413 = 0.1587

            b.         0.5403, first, the area between 18 and 20 is 0.1554.  The area between 20 and 26 is 0.3849.  0.1554 + 0.3849 = 0.5403

            c.         About 1 person, found by z = (7 – 20)/5 = – 2.6, for which the area is 0.4953.  Then 0.5000 – 0.4953 = 0.0047.  Finally, 200(0.0047) = 0.94, which is about one customer.

 

25.        60.06%, found by (42,000 – 40,000)/5000 = 0.40.  The area under the curve for 0.40 is 0.1554.  similarly, the area between 32,000 and 40,000 is 0.4452.  0.1554 + 0.4452 = 0.6006

 

26.        a.         (860 – 1000)/50 = – 2.8.  The area below 860 is 0.0026, found by 0.5000 – 0.4974, or 0.26 percent

            b.         (1055 – 1000)/50 = 1.1.  The area between 1000 and 1055 is 0.3643.  The area between 1000 and 1100 is 0.4772.  Subtracting:  0.4772 – 0.3643 = 0.1129 or 11.29%.

 

27.        About 4099 units found by solving for X.  1.65 = (X – 4000)/60

 

28.        a.         About 0.47% (65,200 – 60,000)/2000 = 2.6.  Then 0.5000 – 0.4953 = 0.0047

            b.         About 22 trucks, (55,000 – 60,000)/2000 = – 2.5.  Then 0.5000 – 0.4938 = 0.0062.  Multiplying, 0.0062 x 3500 = 21.7

            c.         About 2945, (62,000 – 60,000)/2000 = 1.00.  then 0.5000 + 0.3413 = 0.8413.  Multiplying 0.8413 x 3500 = 2944.55

29.        a.         Only 2.28% earn more than John:  (30,400 – 28,000)/1200 = 2.00.  Then 0.5000 – 0.4772 = 0.0228

            b.         Of the other supervisors, 97.72% have more service.  (10 – 20)/5 = – 2.00.  Then 0.4772 + 0.5000 = 0.9772

 

30.        a.         26.43%, found by (30 – 35)/8 = – 0.63, Then 0.500 – 0.2357 = 0.2643

            b.         26.43%, found by (40 – 35)/8 = 0.63.  Then 0.5000 – 0.2357 = 0.2643

            c.         The normal distribution is continuous.  Thus, the probability of an exact value is very small.

            d.         About 4.26%.  You could find the probability of 39.5 and 40.5

                        z = (39.5 – 35)/8 = 0.56.  Area is 0.2123

                        z = (40.5 – 35)/8 = 0.69.  Area is 0.2549

Subtracting:  0.2549 – 0.2123 = 0.0426

            e.         45.24 minutes or longer, found by solving for X.  1.28 = (X – 35)/8


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