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Solutions to Worksheet #2
Finding Probabilities
Standard Normal
x
1.3
−0.8
x
−1.9
−0.8
P[Z<x]
0.9032
0.2119
y
−0.7
1.5
P[Z>x]
0.9713
0.7881
P[Z>x]
0.0968
0.7881
P[Z<y]
0.2420
0.9332
P[x<Z<y]
0.2132
0.7213
Normal N(-0.3,0.36) (¾ = 0:6)
x
0.8
−1
x
0
−0.4
P[X<x]
0.9666
0.1217
y
1.8
0.2
P[X>x]
0.3085
0.5662
P[X>x]
0.0334
0.8783
P[X<y]
0.9998
0.7977
Finding Values
Standard Normal
®
0.05
0.025
0.005
z®
1.645
1.960
2.576
P[x<X<y]
0.3083
0.3639
z-score x
1.83
−1.17
z-score x
0.50
−0.17
z-score y
3.50
0.83
Endpoints for symmetric intervals of given probability p ( ¡z (1¡p)=2 ∙ Z ∙ z (1¡p)=2)
probability p
endpoints z (1¡p)=2
0.8
0.9
0.95
0.99
1.282
1.645
1.960
2.576
Normal N(-0.3,0.36) (¾ = 0:6)
probability p
0.1
zp
1.282
z 1¡p
−1.282
c: X>c
0.469
c:X<c
−1.069
Endpoints for intervals around the mean of probability p (¹ ¡ c ∙ X ∙ ¹ + c)
probability p
z (1¡p)=2
c
0.85
0.92
0.98
1.440
1.751
2.326
0.864
1.050
1.396
¹¡c
−1.164
−1.350
−1.696
¹+c
0.564
0.750
1.096
Additional Problems
Find ¾ such that a normal variable Y » N (9; ¾) is such that P [ j Y ¡ 9 j ∙ 0:05] = 0:99.
A standard normal variable has probability 0.99 to fall between ¡2:576 and 2:576 (the z-score
z 0:005). Hence, Y has probability 0.99 of falling within 2.576 times its standard deviation from its
0:05 ≈ 0:0194,
mean. In other words, we need ¾ to be such that 0:05 = 2:576¾, that is ¾ = 2:576
2
corresponding to a variance ¾ 2 ≈ 0:0194 ≈ 0:0003767
Suppose we are sampling from a N (0; 0:16) distribution. How large should n (the sample size)
¹ j ∙ 0:01] = 0:99?
be so that P [ j X
We have that ¾ = 0:4 For a sample size of n, the sample mean will have expectation 0, and
0:4
variance 0:16
n , or a standard deviation of p n. We already noted that the appropriate z-score for
an interval of probability 0.99 is 2.576. Hence, similarly to the previous question, we need that
µ
¶2
2
0:4
0:4
¢
2:576
p n ¢ 2:576 = 0:01. Solving for n, n =
= (2:576 ¢ 40) ≈ 10;617
¡2
10