oakleylynn
28.06.2020 •
Mathematics
Find the following probability for the standard normal random variable z:
A) P(-1 ≤ z ≤ 1)
B) P(-2 ≤ z ≤ 2)
C) P(-2.61 ≤ z ≤ 0.57)
D) P(-0.95< z < 1.11)
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Ответ:
i) P(-1 ≤ z ≤ 1) = 0.6826
ii) P(-2 ≤ z ≤ 2) = 0.9544
iii) P(-2.61 ≤ z ≤ 0.57) = 0.7112
iv) P(-0.95< z < 1.11) = 0.6954
Step-by-step explanation:
i)
P(-1 ≤ z ≤ 1) = P( z ≤ 1 ) - P(z≤-1)
= 0.5 + A(1) - (0.5 - A(-1))
= 0.5 + A(1) - 0.5 + A(-1)
= A(1) + A(1) (∵ A(-1) = A(1)
= 2 × A(1)
= 2×0.3413 ( From Normal table)
P(-1 ≤ z ≤ 1) = 0.6826
ii)
P(-2 ≤ z ≤ 2) = P( z ≤ 2 ) - P(z≤-2)
= 0.5 + A(2) - (0.5 - A(-2))
= 0.5 + A(2) - 0.5 + A(-2)
= A(2) + A(2) (∵ A(-2) = A(2)
= 2 × A(2)
= 2 × 0.4772
= 0.9544
P(-2 ≤ z ≤ 2) = 0.9544
iii)
P(-2.61 ≤ z ≤ 0.57) = P( z ≤ 0.57)-P(z ≤-2.61)
= 0.5 + A(0.57) - (0.5 - A(-2.61))
= A( 0.57) + A( 2.61) ( ∵ A(-2.61) = A(2.61)
= 0.2157 +0.4955
= 0.7112
P(-2.61 ≤ z ≤ 0.57) = 0.7112
iv)
P(-0.95< z < 1.11) = P( z ≤ 1.11)-P(z ≤-0.95)
= 0.5 + A(1.11) - (0.5 - A(-0.95))
= A( 1.11) + A( 0.95) (∵ A(-0.95) = A(0.95)
= 0.3665 + 0.3289
= 0.6954
P(-0.95< z < 1.11) = 0.6954
Ответ:
Hi there!
~
You have to group by 4.
Hope this helped you!