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6. A researcher develops a test for identifying intellectually gifted children. It has au= 56 and o=8.d.What percentage of children are expected to score above 60?What percentage of the scores will be below 54?b.You measure 64 children and get a sample mean of 58.28? How likely is this sample toC.relate to the originally described population using a two tail analysis with a 0.05. Wouldyour conclusion change if we did a one tail analysis with a-0.01?

Question
6. A researcher develops a test for identifying intellectually gifted children. It has au= 56 and o=8.
d.
What percentage of children are expected to score above 60?
What percentage of the scores will be below 54?
b.
You measure 64 children and get a sample mean of 58.28? How likely is this sample to
C.
relate to the originally described population using a two tail analysis with a 0.05. Would
your conclusion change if we did a one tail analysis with a-0.01?
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6. A researcher develops a test for identifying intellectually gifted children. It has au= 56 and o=8. d. What percentage of children are expected to score above 60? What percentage of the scores will be below 54? b. You measure 64 children and get a sample mean of 58.28? How likely is this sample to C. relate to the originally described population using a two tail analysis with a 0.05. Would your conclusion change if we did a one tail analysis with a-0.01?

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Step 1

Area under the normal curve is the sum of all probabilities of variable X. Since probability ranges from 0 to 1, the total area under the cuve is equal to 1. We can calculate the probability of finding a score/number less than X ie P(x<X)  by calculating Z score and the finding corresponding probability of that Z score using Z table.

But if we need to find probability of a score greater than X ie P(x>X) then we first calculated P(x<X) and subtract that from 1.

Mu-56
60
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Mu-56 60

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Step 2

Part a)

Z score computation:

Z=X-ju
60 56
4
=0.5
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Z=X-ju 60 56 4 =0.5

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Step 3

Probability corresponding to Z score of 0.5 can be read from z table as shown below.

P(X<60)=P(Z<0.5) = ...

00
.06
09
.01
.02
.03
04
.05
.07
.08
Z
5359
5080
5120
5160
5199
5239
5279
5319
0.0
5000
5040
5675
.5714
5753
0.1
5398
.5438
5478
.5517
5557
.5596
5636
5987
.6026
.5832
.5871
.5910
5948
6064
6103
6141
0.2
.5793
6480
6293
6443
6517
0.3
6179
6217
6255
6331
6368
6406
6736
6808
6554
6591
.6628
.6664
6700
.6772
.6844
6879
0.4
7054
7123
7157
7190
7224
0.5
6915
6950
.6985
7019
7088
7389
7517
7549
0.6
7257
7291
7324
7357
7422
.7454
7486
7704
7734
7764
.7794
7823
7852
0.7
7580
7611
7642
7673
0.8
7939
7881
7910
7967
7995
.8023
.8051
.8078
8106
.8133
8186
8212
.8238
8264
8289
8315
8340
.8365
.8389
0.9
8159
.8461
8577
.8621
1.0
.8413
8438
.8485
.8508
.8531
8554
.8599
.8686
8810
.8708
8729
.8749
8770
8790
.8830
1.1
8643
8665
.8997
.9015
1.2
.8849
.8869
.8907
.8925
8944
.8962
8888
8980
9082
9236
9049
9066
.9099
9115
9131
9147
9162
9177
1.3
9032
.9292
.9306
.9319
9192
.9207
.9222
9251
.9265
9279
1.4
0222
024E
0257
0270
0200
0204
0406
0410
0420
0411
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00 .06 09 .01 .02 .03 04 .05 .07 .08 Z 5359 5080 5120 5160 5199 5239 5279 5319 0.0 5000 5040 5675 .5714 5753 0.1 5398 .5438 5478 .5517 5557 .5596 5636 5987 .6026 .5832 .5871 .5910 5948 6064 6103 6141 0.2 .5793 6480 6293 6443 6517 0.3 6179 6217 6255 6331 6368 6406 6736 6808 6554 6591 .6628 .6664 6700 .6772 .6844 6879 0.4 7054 7123 7157 7190 7224 0.5 6915 6950 .6985 7019 7088 7389 7517 7549 0.6 7257 7291 7324 7357 7422 .7454 7486 7704 7734 7764 .7794 7823 7852 0.7 7580 7611 7642 7673 0.8 7939 7881 7910 7967 7995 .8023 .8051 .8078 8106 .8133 8186 8212 .8238 8264 8289 8315 8340 .8365 .8389 0.9 8159 .8461 8577 .8621 1.0 .8413 8438 .8485 .8508 .8531 8554 .8599 .8686 8810 .8708 8729 .8749 8770 8790 .8830 1.1 8643 8665 .8997 .9015 1.2 .8849 .8869 .8907 .8925 8944 .8962 8888 8980 9082 9236 9049 9066 .9099 9115 9131 9147 9162 9177 1.3 9032 .9292 .9306 .9319 9192 .9207 .9222 9251 .9265 9279 1.4 0222 024E 0257 0270 0200 0204 0406 0410 0420 0411

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