Question 1192959: Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weight of coal loaded into each car is normally distributed, with mean of 75 tons and standard deviation of 0.8 ton.
(d) Among 20 randomly chosen cars, most likely, how many cars will be loaded with less than 74.4 tons of coal? Calculate the corresponding probability.
(e) In the senior management meeting, it is discussed and agreed that a car loaded with less than 74.4 tons of coal is not cost effective. To reduce the ratio of cars to be loaded with less than 74.4 tons of coal, it is suggested changing current average loading of coal from 75 tons to a new average level, M tons. Should the new level M be (I) higher than 75 tons or (II) lower than 75 tons? (Write down your suggestion, no explanation is needed in part (e)).
The entries in Table I are the probabilities that a random variable having the standard normal distribution will take on a value between 0 andz. They are given by the area of the gray region under the curve in the figure.
TABLE I
0.01 0.02
0.0040 0.0080 0.0438 0.0478 0.0832 0.0871 0.1217 0.1255 0.1591 0.1628 0.1950 0.1985
0.2291 0.2324 0.2611 0.2642 0.2910 0.2939 0.3186 0.3212 0.3438 0.3461
0.3665 0.3686 0.3869 0.3888 0.4049 0.4066 0.4207 0.4222 0.4345 0.4357
0.4463 0.4474 0.4564 0.4573 0.4648 0.4656 0.4719 0.4725 0.4778 0.4783
0.4826 0.4830 0.4864 0.4868 0.4896 0.4898 0.4920 0.4922 0.4940 0.4941
0.4955 0.4956 0.4966 0.4967 0.4975 0.4976 0.4982 0.4982 0.4987 0.4987
NORMAL-CURVE AREAS
z 0.00
0.0 0.0000
0.1 0.0398
0.2 0.0793
0.3 0.1179
0.4 0.1554
0.5 0.1915
0.6 0.2257
0.7 0.2580
0.8 0.2881
0.9 0.3159
1.0 0.3413
1.1 0.3643
1.2 0.3849
1.3 0.4032
1.4 0.4192
1.5 0.4332
1.6 0.4452
1.7 0.4554
1.8 0.4641
1.9 0.4713
2.0 0.4772
2.1 0.4821
2.2 0.4861
2.3 0.4893
2.4 0.4918
2.5 0.4938
2.6 0.4953
2.7 0.4965
2.8 0.4974
2.9 0.4981
3.0 0.4987
0.03 0.04
0.0120 0.0160 0.0517 0.0557 0.0910 0.0948 0.1293 0.1331 0.1664 0.1700 0.2019 0.2054
0.2357 0.2389 0.2673 0.2704 0.2967 0.2995 0.3238 0.3264 0.3485 0.3508
0.3708 0.3729 0.3907 0.3925 0.4082 0.4099 0.4236 0.4251 0.4370 0.4382
0.4484 0.4495 0.4582 0.4591 0.4664 0.4671 0.4732 0.4738 0.4788 0.4793
0.4834 0.4838 0.4871 0.4875 0.4901 0.4904 0.4925 0.4927 0.4943 0.4945
0.4957 0.4959 0.4968 0.4969 0.4977 0.4977 0.4983 0.4984 0.4988 0.4988
0.05 0.06 0.07 0.08
0.0199 0.0239 0.0279 0.0319 0.0596 0.0636 0.0675 0.0714 0.0987 0.1026 0.1064 0.1103 0.1368 0.1406 0.1443 0.1480 0.1736 0.1772 0.1808 0.1844 0.2088 0.2123 0.2157 0.2190
0.2422 0.2454 0.2486 0.2517 0.2734 0.2764 0.2794 0.2823 0.3023 0.3051 0.3078 0.3106 0.3289 0.3315 0.3340 0.3365 0.3531 0.3554 0.3577 0.3599
0.3749 0.3770 0.3790 0.3810 0.3944 0.3962 0.3980 0.3997 0.4115 0.4131 0.4147 0.4162 0.4265 0.4279 0.4292 0.4306 0.4394 0.4406 0.4418 0.4429
0.4505 0.4515 0.4525 0.4535 0.4599 0.4608 0.4616 0.4625 0.4678 0.4685 0.4692 0.4699 0.4744 0.4750 0.4756 0.4761 0.4798 0.4803 0.4808 0.4812
0.4842 0.4846 0.4850 0.4854 0.4878 0.4881 0.4884 0.4887 0.4906 0.4909 0.4911 0.4913 0.4929 0.4931 0.4932 0.4934 0.4946 0.4948 0.4949 0.4951
0.4960 0.4961 0.4962 0.4963 0.4970 0.4971 0.4972 0.4973 0.4978 0.4979 0.4979 0.4980 0.4984 0.4985 0.4985 0.4986 0.4989 0.4989 0.4989 0.4990
0.09
0.0359 0.0753 0.1141 0.1517 0.1879 0.2224
0.2549 0.2852 0.3133 0.3389 0.3621
0.3830 0.4015 0.4177 0.4319 0.4441
0.4545 0.4633 0.4706 0.4767 0.4817
0.4857 0.4890 0.4916 0.4936 0.4952
0.4964 0.4974 0.4981 0.4986 0.4990
Also, for z = 4.0, 5.0 and 6.0, the
areas are 0.49997, 0.4999997, and 0.499999999.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **d) Find the number of cars with less than 74.4 tons of coal:**
* **1. Standardize the value:**
* We need to find the z-score for 74.4 tons:
* z = (X - μ) / σ
* z = (74.4 - 75) / 0.8
* z = -0.6 / 0.8 = -0.75
* **2. Find the probability of a car having less than 74.4 tons:**
* From the standard normal distribution table (Table I), the area to the left of z = -0.75 is 1 - 0.2734 = 0.7266
* (Since the table gives the area between 0 and z, we need to subtract the table value from 0.5 to get the area to the left of -0.75)
* **3. Find the expected number of cars with less than 74.4 tons:**
* Multiply the probability by the number of cars:
* Expected number of cars = 0.7266 * 20 = 14.532
* **Most likely, 14 or 15 cars will be loaded with less than 74.4 tons of coal.**
**e) Suggestion for the new average loading level (M):**
* **(I) Higher than 75 tons**
By increasing the average loading weight (M), the distribution will shift to the right. This will reduce the proportion of cars loaded with less than 74.4 tons of coal.
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