number 3 ,4 ,5 TABLE D.1AREAS UNDER THE STANDARDIZED NORMAL DISTRIBUTIONExamplePr(0sZs1.96) = 0.4750Pr(Zz1.96) = 0.5 – 0.4750 = 0.0250.47501.96.00.01.02.03.04.05.06.07.08.09.0000.0398.0080.03190.00.1.0040.0438.0120.0160.0199.0239.0279.0359.0478.0517.0557.0596.0636.0675.0714.07530.2.0793.0832.0871.0910.0948.0987.1026.1064.1103.11410.3.1179.1217.1255.1293.1664.1331.1700.1368.1406.1772.1443.1480.15170.40.5.18792224.1554.1591.1628.1736.1808.1844.1915.1950.1985.2019.2054.20882123.21572190225725800.62291.232423572389.2422245424862517.25490.7.2611.2642.2673.2704.27342764.2794282328520.8.2881.2910.2939.296729953023.3051.3078.3106.3133.3159.33150.91.0.3186.3212.3461.3238.3485.3264.3508.3289.3531.3340.3577.3365.3599.3389.3621.3413.3438.3554.3708.39071.1.36433830.3665.3869.3686.3729.3749.3770.3790.38101.2.3849.3888.3925.3944.3962.39803997.40151.34032.4049.4066.4082.4099.4115.4131.4147.4162417742791.41.5.4192.4207.4222.4236.4251.42654292430643194332.4345.4357.4370.4382.43944406.44184429.4441.4495.45154535.46251.6.4452.4454.4463.4564.4474.4573.4484.4582.4505.4525.4616.4545.46331.7.4591.4599.46081.8.4641.4649.4656.4664.4671.4678.4686.469347564699.47061.9.4713.4719.4726.4732.4738.4744.4750.476147672.0.4772.4778.4783.4788.4793.4798.4803.4808.481248172.1.4821.4826.48304834.4838.4842.4846.4850.485448574890.4916.4868.4871.4901.4878.4906.4881.488749132.2.4861.4893.4864.487548842.3.4896.4898.4904.4909.49112.42.5.49184938.4920.4922.4941.4925.4927.4945.4929.4946.49314948.4932.4949.4934.49514936.4952.4940.494349612.62.7.4953.4965.4955.4966.4956.4957.4959.4960.4962.4963.4964.4967.4968.4969.4970.49714972497349742.8.4974.4975.4976.4977.4977.49784979.4979498049812.93.0.4981.4987.4982.49874982.4983.4984.4984.498549854986.4986.4987.4988.4988.4989.4989.498949904990Note: This table gives the area in the right-hand tail of the distribution (i.e., Zz 0). But since the normaldistribution is symmetrical about 2=0, the area in the left-hand tail is the same as the area in the corrospondingright-hand tail. For example, P(-1.96 s Z<0) = 0.4750. Thorofore, P-1.96 sZs 1.96) = 2(0.4750) = 0.95.6789 oO O O 0002222N2222 N3 Question FiveThe management of a supermarket wants to adopt a new promotional policy ofgiving free gift to every customer who spends more than a certain amount per visitat this supermarket. The expectation of the management is that after thispromotional policy is advertised, the expenditure for all customers at thissupermarket 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 freegift?2) In a sample of 100 customers, what are the number of customers whoseexpenditure is between 420 £ and 485 £?3) What is a probability of selecting a customer whose expenditure is differ thanthe population mean expenditure by at most 50 £?4) In a sample of 49 customers, what are the number of customers whosemean expenditure is at least 410 £?5) What is the probability that the expenditure of the first customer exceeds theexpenditure of the second customer by at least 20 £?

Name: number 3 ,4 ,5 TABLE D.1AREAS UNDER THE STANDARDIZED NORMAL DISTRIBUTI
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Description: number 3 ,4 ,5 TABLE D.1AREAS UNDER THE STANDARDIZED NORMAL DISTRIBUTIONExamplePr(0sZs1.96) = 0.4750Pr(Zz1.96) = 0.5 – 0.4750 = 0.0250.47501.96.00.01.02.03.04.05.06.07.08.09.0000.0398.0080.03190.00.1.0040.0438.0120.0160.0199.0239.0279.0359.0478.0517.

Given Information: The expenditure for all customers at this supermarket is normally distributed…

Math

Statistics

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

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

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.6: Summarizing Categorical Data

Problem 23PPS

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Question

number 3 ,4 ,5

TABLE D.1
AREAS UNDER THE STANDARDIZED NORMAL DISTRIBUTION
Example
Pr(0sZs1.96) = 0.4750
Pr(Zz1.96) = 0.5 – 0.4750 = 0.025
0.4750
1.96
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0000
.0398
.0080
.0319
0.0
0.1
.0040
.0438
.0120
.0160
.0199
.0239
.0279
.0359
.0478
.0517
.0557
.0596
.0636
.0675
.0714
.0753
0.2
.0793
.0832
.0871
.0910
.0948
.0987
.1026
.1064
.1103
.1141
0.3
.1179
.1217
.1255
.1293
.1664
.1331
.1700
.1368
.1406
.1772
.1443
.1480
.1517
0.4
0.5
.1879
2224
.1554
.1591
.1628
.1736
.1808
.1844
.1915
.1950
.1985
.2019
.2054
.2088
2123
.2157
2190
2257
2580
0.6
2291
.2324
2357
2389
.2422
2454
2486
2517
.2549
0.7
.2611
.2642
.2673
.2704
.2734
2764
.2794
2823
2852
0.8
.2881
.2910
.2939
.2967
2995
3023
.3051
.3078
.3106
.3133
.3159
.3315
0.9
1.0
.3186
.3212
.3461
.3238
.3485
.3264
.3508
.3289
.3531
.3340
.3577
.3365
.3599
.3389
.3621
.3413
.3438
.3554
.3708
.3907
1.1
.3643
3830
.3665
.3869
.3686
.3729
.3749
.3770
.3790
.3810
1.2
.3849
.3888
.3925
.3944
.3962
.3980
3997
.4015
1.3
4032
.4049
.4066
.4082
.4099
.4115
.4131
.4147
.4162
4177
4279
1.4
1.5
.4192
.4207
.4222
.4236
.4251
.4265
4292
4306
4319
4332
.4345
.4357
.4370
.4382
.4394
4406
.4418
4429
.4441
.4495
.4515
4535
.4625
1.6
.4452
.4454
.4463
.4564
.4474
.4573
.4484
.4582
.4505
.4525
.4616
.4545
.4633
1.7
.4591
.4599
.4608
1.8
.4641
.4649
.4656
.4664
.4671
.4678
.4686
.4693
4756
4699
.4706
1.9
.4713
.4719
.4726
.4732
.4738
.4744
.4750
.4761
4767
2.0
.4772
.4778
.4783
.4788
.4793
.4798
.4803
.4808
.4812
4817
2.1
.4821
.4826
.4830
4834
.4838
.4842
.4846
.4850
.4854
4857
4890
.4916
.4868
.4871
.4901
.4878
.4906
.4881
.4887
4913
2.2
.4861
.4893
.4864
.4875
4884
2.3
.4896
.4898
.4904
.4909
.4911
2.4
2.5
.4918
4938
.4920
.4922
.4941
.4925
.4927
.4945
.4929
.4946
.4931
4948
.4932
.4949
.4934
.4951
4936
.4952
.4940
.4943
4961
2.6
2.7
.4953
.4965
.4955
.4966
.4956
.4957
.4959
.4960
.4962
.4963
.4964
.4967
.4968
.4969
.4970
.4971
4972
4973
4974
2.8
.4974
.4975
.4976
.4977
.4977
.4978
4979
.4979
4980
4981
2.9
3.0
.4981
.4987
.4982
.4987
4982
.4983
.4984
.4984
.4985
4985
4986
.4986
.4987
.4988
.4988
.4989
.4989
.4989
4990
4990
Note: This table gives the area in the right-hand tail of the distribution (i.e., Zz 0). But since the normal
distribution is symmetrical about 2=0, the area in the left-hand tail is the same as the area in the corrosponding
right-hand tail. For example, P(-1.96 s Z<0) = 0.4750. Thorofore, P-1.96 sZs 1.96) = 2(0.4750) = 0.95.
6789 o
O O O 000
2222N2
222 N3

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 £?

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