Respuesta :
Answer:
a)
The probability that a single score drawn at random will be greater than 110
P( X > 110) = 0.1587
b)
The probability that a sample of 25 scores will have a mean greater than 105
P( x> 105) = 0.0062
c)
The probability that a sample of 64 scores will have a mean greater than 105
P( x⁻> 105) = 0.002
d)
The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105
P( 95 ≤ X≤ 105) = 0.9544
Step-by-step explanation:
a)
Given mean of the Normal distribution 'μ' = 100
Given standard deviation of the Normal distribution 'σ' = 10
a)
Let 'X' be the random variable of the Normal distribution
let 'X' = 110
[tex]Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1[/tex]
The probability that a single score drawn at random will be greater than 110
P( X > 110) = P( Z >1)
= 1 - P( Z < 1)
= 1 - ( 0.5 +A(1))
= 0.5 - A(1)
= 0.5 -0.3413
= 0.1587
b)
let 'X' = 105
[tex]Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5[/tex]
The probability that a single score drawn at random will be greater than 110
P( x> 105) = P( z > 2.5)
= 1 - P( Z< 2.5)
= 1 - ( 0.5 + A( 2.5))
= 0.5 - A ( 2.5)
= 0.5 - 0.4938
= 0.0062
The probability that a single score drawn at random will be greater than 105
P( x> 105) = 0.0062
c)
let 'X' = 105
[tex]Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4[/tex]
The probability that a single score drawn at random will have a mean greater than 105
P( x> 105) = P( z > 4)
= 1 - P( Z< 4)
= 1 - ( 0.5 + A( 4))
= 0.5 - A ( 4)
= 0.5 - 0.498
= 0.002
The probability that a sample of 64 scores will have a mean greater than 105
P( x⁻> 105) = 0.002
d)
Let x₁ = 95
[tex]Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2[/tex]
Let x₂ = 105
[tex]Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2[/tex]
The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105
P( 95 ≤ X≤ 105) = P( -2≤z≤2)
= P(z≤2) - P(z≤-2)
= 0.5 + A( 2) - ( 0.5 - A( -2))
= A( 2) + A(-2) (∵A(-2) =A(2)
= A( 2) + A(2)
= 2 × A(2)
= 2×0.4772
= 0.9544
The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105
P( 95 ≤ X≤ 105) = 0.9544
