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Standard Normal Distribution Binomial Distribution 21 22 Area to the left of zi Area to the left of z2 Desired Area N p X p(x

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Answer #1

P (    0.200   < Z <    1.500   )
= P ( Z <    1.500   ) - P ( Z <   0.20   ) =
0.9332   -    0.5793   =    0.3539

P (    -1.200   < Z <    0.300   )
= P ( Z <    0.300   ) - P ( Z <   -1.20   ) =
0.6179   -    0.1151   =    0.5028

X P(X)
0 0.0008
1 0.0068
2 0.0278
3 0.0716
4 0.1304
5 0.1789
6 0.1916
7 0.1643
8 0.1144
9 0.0654
10 0.0308
11 0.0120
12 0.0039
13 0.0010
14 0.0002
15 0.0000
16 0.0000
17 0.0000
18 0.0000
19 0.0000
20 0.0000

Mean = np =    20*0.3=       6


Variance = np(1-p) =    20*0.3*(1-0.3)=      4.2

Please let me know in case of any doubt.

Thanks in advance!


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