Transcribed Text
Complete and post the following problems.
1. We have a population that is normally distributed. A sample of 10 gives an
average of 4.00 and a standard deviation of 1.873. What is the probability of
pulling a single sample between 3 and 5? Use the posted Table for the Student
t distribution. The pdf file is located directly beneath the assignment
document. Use that table to solve these two problems. Use linear
interpolation to solve (basic algebra) these two problems.
2. From the above problem; what is the probability of pulling a single sample less
than 1.6?
Table A2 The , distribution
The entry in TOW * (degrees of freedomi
under column beading P (probability) is the
value 1. for which POSISION = e
p
0.05
0.1
0.2
0.25
0.3
0.4
0.45
0.475
0.49
0.495
A
I
158
325
.727
1,000
1.376
3.08
6.31
12.71
31.82
63.66
2
142
289
617
816
1.06)
1.89
2.92
4,30
6.96
9.92
3
137
277
584
765
978
1.64
2.35
3.18
4.54
5.84
4
134
271
569
.741
941
1.53
2.13
27%
3.75
4.60
3
132
267
559
727
920
1.48
2:02
2.57
3.36
4.03
6
131
265
553
318
906
1.44
1.94
2.45
3.14
3.71
7
.130
263
549
711
896
1.42
1.90
2.36
3.00
3.50
8
130
262
.546
706
889
1.40
1.86
2,31
2,90
3.36
9
129
261
543
703
883
1.38
1.83
2.26
1.82
3.25
10
129
260
142
700
879
1.37
1.81
2.23
2.76
3.17
11
.129
260
540
697
876
1.36
1.80
2.20
272
3,11
12
128
259
.539
695
873
1.36
1.78
2.18
2,68
3.06
13
12M
259
538
694
870
1.35
1.77
216
2.65
3.01
14
128
258
537
692
868
1.34
1.76
2.14
2.62
2.98
15
128
258
536
691
866
1.34
1.75
2.13
2.60
2.95
16
128
258
535
690
865
1.34
1.75
1.12
2.58
2.92
17
128
257
534
689
863
1.33
1.74
2.11
2.57
2.90
18
127
297
534
688
862
1.33
1.73
2,10
2.55
2.88
19
127
257
.533
.688
861
1.33
1.73
1.09
2.54
2.86
20
127
257
.533
687
860
1.32
1.72
2.09
2.53
2.84
21
127
257
532
686
859
1.32
1.72
2.08
2.53
2.83
22
127
256
532
686
.558
1.32
1.72
2.07
251
2.62
23
127
256
532
685
858
1.32
1.71
2.07
2.50
2.81
24
127
256
531
685
837
1.32
1.71
2.06
2.49
2,80
25
127
256
531
694
856
1.33
1.71
2.06
2.48
2.79
26
127
256
531
684
856
1.32
1.71
2.06
2.48
2.78
27
127
356
531
684
855
1,31
1.70
2.05
2.47
2.77
28
127
256
530
683
855
1.31
1.70
2.05
2.47
2.76
29
127
256
.330
683
.854
1.31
1.70
204
2.46
2.76
30
127
256
530
683
.854
1.31
1.70
2,04
2.46
2.75
40
126
255
529
681
851
1.30
1.68
2.02
2.42
2.70
60
136
254
527
679
848
1.30
1.67
2.00
2.39
2.66
120
.126
254
536
677
845
1.29
1.66
1.98
2.36
2.62

.126
253
524
674
842
1.28
1.645
1.96
1.33
2.58
Sownvel R. A Fisher and F. Yates, Tollies for anuf Medical published by Longman
Group Ltd., London (previously published by Hiver and Bnyd. and by permission of the authors and
publishen.
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1.
Given that mean = 4.00 and standard deviation (SD) = 1.873
then, for P(3 < x < 5)
t1 =( 3mean)/SD/(sqrt(n)) = (34)/(1.873/sqrt(10)) = 1.69
t2 =( 5mean)/SD/(sqrt(n)) = (54)/(1.873/sqrt(10)) = 1.69
Thus Probability P(3 < x < 5) = P(1.69 < t < 1.69) .
Since t distribution is symmetric, therefore
P(1.69 < t < 1.69) = 2P(0 < t < 1.69)
Also sample size is 10. Thus we will check probability and t value at k = 101 = 9 in the table...