Question

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that a single score drawn at random will be greater than 110 (b) the probability that a sample of 25 scores will have a mean greater than 105 (c) the probability that a sample of 64 scores will have a mean greater than 105 (d) the probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

Answer 1

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

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

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

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

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

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

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

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

Let  x₂ = 105

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

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