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(b) Required area area between z 0 and z 0.68 (by symmetry). Therefore, proceed downward under column marked z until entry 0.6 is reached. Then proceed right to column marked 8.
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CHAPTER 4 Special Probability Distributions
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The result, 0.2517, is the required area and represents the probability that Z is between (Fig. 4-4). Therefore, P( 0.68 Z 0)
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0 1 3 0.68e !2p 0.68 1 30 e !2p u2>2 du
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0.68 and 0
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(c) Required area
(area between z (area between z (area between z (area between z 0.1772 0.4864
0.46 and z 0 and z 0 and z 0 and z 0.6636
0) 2.21)
0.46) 2.21)
The area, 0.6636, represents the probability that Z is between P( 0.46 Z 2.21) 1 22p 1 22p 1 22p
0.46 and 2.21 (Fig. 4-5). Therefore,
u2> 2
du du 1 22p 1 22p
u2/2
30 e
u2>2
du 0.1772 0.4864
30 e
u2/2
30 e
u2/2
0.6636 (d) Required area (Fig. 4-6) (area between z (area between z 0.4738 This is the same as P(0.81 (e) Required area (Fig. 4-7) Z 0.2910 0 and z 0 and z 0.1828 1.94) 0.81)
1.94). 1.28 and z 0) 0.8997 0)
(area between z (area to right of z 0.3997 0.5
This is the same as P(Z
1.28).
Fig. 4-6
Fig. 4-7
4.13. If area refers to that under the standard normal curve, find the value or values of z such that (a) area between 0 and z is 0.3770, (b) area to left of z is 0.8621, (c) area between 1.5 and z is 0.0217.
CHAPTER 4 Special Probability Distributions
(a) In the table in Appendix C the entry 0.3770 is located to the right of the row marked 1.1 and under the column marked 6. Then the required z 1.16. By symmetry, z 1.16 is another value of z. Therefore, z 1.16 (Fig. 4-8). The problem is equivalent to solving for z the equation
z 1 30 e !2p u2>2 du
(b) Since the area is greater than 0.5, z must be positive. Area between 0 and z is 0.8621 0.5 0.3621, from which z
1.09 (Fig. 4-9).
Fig. 4-8
Fig. 4-9
(c) If z were positive, the area would be greater than the area between must be negative. Case 1 z is negative but to the right of Area between 0.0217 1.5 (Fig. 4-10). (area between
1.5 and 0, which is 0.4332; hence z
1.5 and z 0.4332
1.5 and 0)
(area between 0 and z) (area between 0 and z) 0.4115 from which z 1.35.
Then the area between 0 and z is 0.4332
Fig. 4.10
Fig. 4.11
Case 2
z is negative but to the left of
1.5 (Fig. 4-11). 1.5 (area between z and 0) (area between 1.5 and 0) 0.4332 1.694 by using linear interpolation;
Area between z and 0.0217
(area between 0 and z)
Then the area between 0 and z is 0.0217 0.4332 or, with slightly less precision, z 1.69.
0.4549 and z
4.14. The mean weight of 500 male students at a certain college is 151 lb and the standard deviation is 15 lb. Assuming that the weights are normally distributed, find how many students weigh (a) between 120 and 155 lb, (b) more than 185 lb.
(a) Weights recorded as being between 120 and 155 lb can actually have any value from 119.5 to 155.5 lb, assuming they are recorded to the nearest pound (Fig. 4-12). 119.5 lb in standard units 155.5 lb in standard units (119.5 2.10 (155.5 0.30 151)>15 151)>15
CHAPTER 4 Special Probability Distributions
Required proportion of students (area between z (area between z (area between z 0.4821 0.1179 2.10 and z 2.10 and z 0 and z 0.6000 300 0.30) 0) 0.30)
Then the number of students weighing between 120 and 155 lb is 500(0.6000)
Fig. 4-12
Fig. 4-13
(b) Students weighing more than 185 lb must weigh at least 185.5 lb (Fig. 4-13). 185.5 lb in standard units (185.5 151)>15 2.30) 0) 0 and z 2.30) 2.30 Required proportion of students (area to right of z (area to right of z (area between z 0.5 0.4893
Then the number of students weighing more than 185 lb is 500(0.0107) 5. If W denotes the weight of a student chosen at random, we can summarize the above results in terms of probability by writing P(119.5 W 155.5) 0.6000 P(W 185.5) 0.0107
4.15. The mean inside diameter of a sample of 200 washers produced by a machine is 0.502 inches and the standard deviation is 0.005 inches. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 inches, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.
0.496 in standard units 0.508 in standard units Proportion of nondefective washers (area under normal curve between z (twice the area between z 2(0.3849) 0.7698, or 77% 77% 23% (Fig. 4-14). 0 and z 1.2 and z 1.2) 1.2) (0.496 (0.508 0.502)>0.005 0.502)>0.005 1.2 1.2
Therefore, the percentage of defective washers is 100%
Fig. 4-14
Note that if we think of the interval 0.496 to 0.508 inches as actually representing diameters of from 0.4955 to 0.5085 inches, the above result is modified slightly. To two significant figures, however, the results are the same.
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