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3 years ago

Find the following probability for the standard normal random variable z:

Mathematics

1 answer:

quester [9]3 years ago

8 0

Answer:

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:

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

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