Question Five The management of a supermarket wants to adopt a new promotional policy of giving free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditure for all customers at this supermarket will be normally distributed with mean 400 £ and a variance of 900 £?. 1) If the management wants to give free gifts to at most 10% of the customers, what should the amount be above which a customer would receive a free gift? 2) In a sample of 100 customers, what are the number of customers whose expenditure is between 420 £ and 485 £? 3) What is a probability of selecting a customer whose expenditure is differ than the population mean expenditure by at most 50 £? 4) In a sample of 49 customers, what are the number of customers whose mean expenditure is at least 410 £? 5) What is the probability that the expenditure of the first customer exceeds the expenditure of the second customer by at least 20 £?

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Question Five
The management of a supermarket wants to adopt a new promotional policy of
giving free gift to every customer who spends more than a certain amount per visit
at this supermarket. The expectation of the management is that after this
promotional policy is advertised, the expenditure for all customers at this
supermarket will be normally distributed with mean 400 £ and a variance of 900
£?.
1) If the management wants to give free gifts to at most 10% of the customers,
what should the amount be above which a customer would receive a free
gift?
2) In a sample of 100 customers, what are the number of customers whose
expenditure is between 420 £ and 485 £?
3) What is a probability of selecting a customer whose expenditure is differ than
the population mean expenditure by at most 50 £?
4) In a sample of 49 customers, what are the number of customers whose
mean expenditure is at least 410 £?
5) What is the probability that the expenditure of the first customer exceeds the
expenditure of the second customer by at least 20 £?
Transcribed Image Text:Question Five The management of a supermarket wants to adopt a new promotional policy of giving free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditure for all customers at this supermarket will be normally distributed with mean 400 £ and a variance of 900 £?. 1) If the management wants to give free gifts to at most 10% of the customers, what should the amount be above which a customer would receive a free gift? 2) In a sample of 100 customers, what are the number of customers whose expenditure is between 420 £ and 485 £? 3) What is a probability of selecting a customer whose expenditure is differ than the population mean expenditure by at most 50 £? 4) In a sample of 49 customers, what are the number of customers whose mean expenditure is at least 410 £? 5) What is the probability that the expenditure of the first customer exceeds the expenditure of the second customer by at least 20 £?
TABLE D.1
AREAS UNDER THE STANDARDIZED NORMAL DISTRIBUTION
Example
Pr(0sZ<1.96)-0.4750
Pr(Z z1.96) -0.5-0.4750 -0.025
0.4750
1.96
00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
.0000
.0398
.0040
.0080
.0120
.0517
.0160
0199
.0239
.0279
.0675
.0319
.0359
0.1
.0438
.0478
.0596
.0987
.1368
.1736
.0636
.1026
.1406
.1772
2123
.0557
.0714
.0753
0.2
.0793
.0832
.0871
.1255
.0910
.0948
.1331
1700
.1064
.1103
.1480
.1844
2190
.1141
0.3
.1179
.1554
.1915
.1217
1293
.1443
.1808
2157
.1517
0.4
.1591
.1628
.1664
.1879
2224
0.5
.1950
.1985
2019
2054
2088
0.6
2257
2580
2291
2324
2642
2357
2389
2422
2454
2486
2517
2549
2852
0.7
2611
.2673
2704
2734
2764
2794
2823
0.8
2881
2910
2939
2967
3051
3315
3554
2995
.3023
.3078
.3133
3389
3106
0.9
3159
.3413
.3186
.3438
.3212
3238
3264
3289
3340
.3365
1.0
.3461
3485
.3508
3531
3577
3599
3621
1.1
.3643
.3665
.3686
.3708
.3729
.3749
3770
3790
.3810
3830
1.2
.3849
.3869
4049
3888
3907
3925
3944
3962
.3980
4147
.3997
4015
1.3
.4032
4066
.4082
.4099
.4115
4131
4162
4177
1.4
.4192
.4207
4222
4236
4251
4265
4394
4279
4292
4306
4319
1.5
.4332
.4345
4357
.4370
4382
4406
4418
4429
4441
1.6
.4452
4463
4474
.4484
4495
4505
4515
4608
4525
4616
.4535
4625
4545
1.7
.4454
.4564
4582
4664
4573
4591
.4599
4633
1.8
.4649
4719
.4641
4671
4738
4706
4767
4656
4678
4686
4693
.4699
1.9
.4713
4726
.4732
4744
4750
.4756
.4761
2.0
.4772
.4778
.4783
.4788
.4793
.4798
4803
4808
4812
.4817
2.1
.4821
.4826
.4830
.4834
4838
4842
4846
.4850
4854
.4857
2.2
.4861
.4864
4871
4901
4868
4878
4890
4916
4875
4881
4884
4887
2.3
.4893
4896
4898
4904
4913
4934
.4951
4906
4909
.4911
2.4
.4918
.4920
.4940
.4922
4941
4925
4927
4929
.4931
4932
4936
2.5
.4938
4943
.4945
4946
.4948
4949
.4952
.4955
.4966
2.6
.4953
4956
4964
4974
4957
4959
4960
4961
4962
4963
2.7
.4965
.4967
.4968
.4969
4970
4971
4972
4973
2.8
.4974
.4975
4976
4977
4977
4979
.4978
4984
4979
4985
4980
4981
4986
2.9
.4981
4982
.4982
4983
4984
.4985
4986
3.0
.4987
4987
4987
.4988
.4988
4989
4989
4989
4990
A990
Note: This table gives the area in the right-hand tail of the distribution (1.e., Z 0). But since the normal
distribution is symmetrical about Z=0, the area in the left-hand tail is the same as the area in the corresponding
right-hand tail. For example, A-1.96 s Zs 0) = 0.4750. Therefore, P(-1.96 s Zs 1.96)- 2(0.4750) 0.95.
tab =
cs Scanned with CamScanner
Transcribed Image Text:TABLE D.1 AREAS UNDER THE STANDARDIZED NORMAL DISTRIBUTION Example Pr(0sZ<1.96)-0.4750 Pr(Z z1.96) -0.5-0.4750 -0.025 0.4750 1.96 00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .0000 .0398 .0040 .0080 .0120 .0517 .0160 0199 .0239 .0279 .0675 .0319 .0359 0.1 .0438 .0478 .0596 .0987 .1368 .1736 .0636 .1026 .1406 .1772 2123 .0557 .0714 .0753 0.2 .0793 .0832 .0871 .1255 .0910 .0948 .1331 1700 .1064 .1103 .1480 .1844 2190 .1141 0.3 .1179 .1554 .1915 .1217 1293 .1443 .1808 2157 .1517 0.4 .1591 .1628 .1664 .1879 2224 0.5 .1950 .1985 2019 2054 2088 0.6 2257 2580 2291 2324 2642 2357 2389 2422 2454 2486 2517 2549 2852 0.7 2611 .2673 2704 2734 2764 2794 2823 0.8 2881 2910 2939 2967 3051 3315 3554 2995 .3023 .3078 .3133 3389 3106 0.9 3159 .3413 .3186 .3438 .3212 3238 3264 3289 3340 .3365 1.0 .3461 3485 .3508 3531 3577 3599 3621 1.1 .3643 .3665 .3686 .3708 .3729 .3749 3770 3790 .3810 3830 1.2 .3849 .3869 4049 3888 3907 3925 3944 3962 .3980 4147 .3997 4015 1.3 .4032 4066 .4082 .4099 .4115 4131 4162 4177 1.4 .4192 .4207 4222 4236 4251 4265 4394 4279 4292 4306 4319 1.5 .4332 .4345 4357 .4370 4382 4406 4418 4429 4441 1.6 .4452 4463 4474 .4484 4495 4505 4515 4608 4525 4616 .4535 4625 4545 1.7 .4454 .4564 4582 4664 4573 4591 .4599 4633 1.8 .4649 4719 .4641 4671 4738 4706 4767 4656 4678 4686 4693 .4699 1.9 .4713 4726 .4732 4744 4750 .4756 .4761 2.0 .4772 .4778 .4783 .4788 .4793 .4798 4803 4808 4812 .4817 2.1 .4821 .4826 .4830 .4834 4838 4842 4846 .4850 4854 .4857 2.2 .4861 .4864 4871 4901 4868 4878 4890 4916 4875 4881 4884 4887 2.3 .4893 4896 4898 4904 4913 4934 .4951 4906 4909 .4911 2.4 .4918 .4920 .4940 .4922 4941 4925 4927 4929 .4931 4932 4936 2.5 .4938 4943 .4945 4946 .4948 4949 .4952 .4955 .4966 2.6 .4953 4956 4964 4974 4957 4959 4960 4961 4962 4963 2.7 .4965 .4967 .4968 .4969 4970 4971 4972 4973 2.8 .4974 .4975 4976 4977 4977 4979 .4978 4984 4979 4985 4980 4981 4986 2.9 .4981 4982 .4982 4983 4984 .4985 4986 3.0 .4987 4987 4987 .4988 .4988 4989 4989 4989 4990 A990 Note: This table gives the area in the right-hand tail of the distribution (1.e., Z 0). But since the normal distribution is symmetrical about Z=0, the area in the left-hand tail is the same as the area in the corresponding right-hand tail. For example, A-1.96 s Zs 0) = 0.4750. Therefore, P(-1.96 s Zs 1.96)- 2(0.4750) 0.95. tab = cs Scanned with CamScanner
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