using the table given of normal distribution answer these questions. The time taken by employees at Grace Floral shop to put a bouquet together has a normal distribution with mean 26.4 minutes and standard deviation .8 minutes.(i) Find the probability that an employee chosen at random takes between 24.6 and 27.8 minutes to put a bouquet together. (ii) 12% of employees take more than t minutes to put a bouquet together. (ii) Find the value of t. Table 3AREAS IN TAIL OF THE NORMAL DISTRIBUTIONThe function tabulated is 1 - +(u) where (u) is the cumulative distribution function of astandardised Normal variable u. Thus 1 - (u)= S e-u/2 du is the probability that av2nstanda rdised Normal variable selected at random will be greater(-=)1- (u)than a value of u(x - ). 03. 04.00.01.02.05.06.07.0809. 5000.4602.4207.3821.3446.4960.4562.4168.3783.3409.4920.4522.4129.3745.3372.4880.4483:4090.3707.3336.4840.4443.4052.3669.3300.4801.4404.4013.3632.32640.0.4761.43643974.35943228.4721.43250.10.20.30.4.3936.3557.3192.4681.4286.3897.3520.3156.4641.4247.3859.3483.31210.50.6.3085.2743.2420.2119.3050.2709.2389.2090.3015.2676.2358.2061.1788.2981.2643.2327.2033.2946.2611.2296.2005.17362912.2578.2266.1977.1711.2877.2546.2236.2843.2514.2206.1922.2810.2483.2177.1894.2776.2451.2148. 1867. 16110.70.8.1949.16850.9.1841.1814.1762.1660.1635. 1587. 1357.1151. 0968.0808.1562. 1335.1131.1539.1314.1515.1292.1093.0918.0764.1446. 1230.1038.0869.0721.1423. 1210.10201.0.1492. 1271.1075.0901.1469.1251.1056.1401.1190.1003.0838.0694.13791.11.21.31.4.1112.0934.0778.1170. 0985.0823.0681.0951.0793.0853.0708.0885.0749.07351.51.61.71.81.9.0668.0548.0446.0359.02 87.0655.0537.0436.0351.02 81.0643.0526.0427.0344. 0274.0630.0516.0418.0336.0268.0618.0505.0409.0329.0262.0606.0495..0401.0322. 0256.0594.0485.0392. 0314.0250.0582.0475..0384.0307.0244.0571.0465.0375.0301.0559.0455.0367.0294.0233.0239. 02275.01786 .01743 .01700.01390 .01355 .01321.01072 .01044 .01017.00820 - .00798 .00776. 02222 .021692.02.12.22.302118.01659.01287. 02068 . 02018 .01970 .01923.01539 .01500.01191. 00939 . 00914.00695.01876 .0183101463.01618 .01578.01255 .01222. 00964. 00734 .00714.01160.00889.00676.01426.01130 .0110100866 .00842.00657.009902.4.00755. 00639. 00621.00466.00347.00256.00187. 005232.52.62.72.8.00604 .00587.00440.00336 .00326. 00240.00175. 00508.00379.002 80. 002 05.00149.00570.00539.00427 .00415 .00402 .00391. 00307 .00298.00226 .00219.00164 . 00159.00554.00494.00368 . 00357.00272 .00264.00199 .00193.00144 .00139.0048000453.002 89.00212.00154.00317.00248.00181.00233.001692.93.0.00135.00097.000693.13.23.33.4.00048.000343.53.63.7.00023. 00016.000113.83.9.00007.000054.000003

Given information- Population mean, µ = 26.4 minutes Population standard deviation, σ = 0.8 minutes…

Answered: The time taken by employees at Grace…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

using the table given of normal distribution answer these questions.

The time taken by employees at Grace Floral shop to put a bouquet together has a normal distribution with mean 26.4 minutes and standard deviation .8 minutes.

(i) Find the probability that an employee chosen at random takes between 24.6 and 27.8 minutes to put a bouquet together.

(ii) 12% of employees take more than t minutes to put a bouquet together. (ii) Find the value of t.

Table 3
AREAS IN TAIL OF THE NORMAL DISTRIBUTION
The function tabulated is 1 - +(u) where (u) is the cumulative distribution function of a
standardised Normal variable u. Thus 1 - (u)= S e-u/2 du is the probability that a
v2n
standa rdised Normal variable selected at random will be greater
(-=)
1- (u)
than a value of u
(x - )
. 03
. 04
.00
.01
.02
.05
.06
.07
.08
09
. 5000
.4602
.4207
.3821
.3446
.4960
.4562
.4168
.3783
.3409
.4920
.4522
.4129
.3745
.3372
.4880
.4483
:4090
.3707
.3336
.4840
.4443
.4052
.3669
.3300
.4801
.4404
.4013
.3632
.3264
0.0
.4761
.4364
3974
.3594
3228
.4721
.4325
0.1
0.2
0.3
0.4
.3936
.3557
.3192
.4681
.4286
.3897
.3520
.3156
.4641
.4247
.3859
.3483
.3121
0.5
0.6
.3085
.2743
.2420
.2119
.3050
.2709
.2389
.2090
.3015
.2676
.2358
.2061
.1788
.2981
.2643
.2327
.2033
.2946
.2611
.2296
.2005
.1736
2912
.2578
.2266
.1977
.1711
.2877
.2546
.2236
.2843
.2514
.2206
.1922
.2810
.2483
.2177
.1894
.2776
.2451
.2148
. 1867
. 1611
0.7
0.8
.1949
.1685
0.9
.1841
.1814
.1762
.1660
.1635
. 1587
. 1357
.1151
. 0968
.0808
.1562
. 1335
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.0838
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1.1
1.2
1.3
1.4
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.0778
.1170
. 0985
.0823
.0681
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.0793
.0853
.0708
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1.5
1.6
1.7
1.8
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.02 87
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.0537
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.01786 .01743 .01700
.01390 .01355 .01321
.01072 .01044 .01017
.00820 - .00798 .00776
. 02222 .02169
2.0
2.1
2.2
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02118
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.01287
. 02068 . 02018 .01970 .01923
.01539 .01500
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. 00939 . 00914
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00866 .00842
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Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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