Question

30. Suppose that values are repeatedly chosen from a standard nomal distribution a. In the long run, what proportion of values will be b. What is the long run proportion of selected values e. What is the long-run proportion of values that will d. What is the long-run proportion of values that will e. In the longnun, what proportion of selected values at most 2.15? Less than 2.15 that will exceed 1.50? That will exceed-2.00? be between-1.23 and 2.85 exceed 5? That will exceed-5? z will satisfy 12.507

Answer 1

The values are choosen from standard normal distribution.

Standard normal distribution has mean = $\mu$ = 0 and standard deviation = $\sigma$ = 1

a )

we have to find P( z <= 2.15 )

Using Excel function, { =NORMSDIST( z )    This function returns area less than z }

P( z <= 2.15 ) = NORMSDIST(2.15 ) = 0.9842

Proportion of the values at most 2.15 is the same as the proportion of values less than 2.15 = 0.9842

b )

i ) We have to find P( Z > 1.50 )

P( Z > 1.50 ) = 1 - P( Z <= 1.50 )

Using Excel, P( z <= 1.50) = NORMSDIST(1.50) = 0.9332

P( Z > 1.50 ) = 1 - 0.9332 = 0.0668

Proportion of the values will be exceeds 1.50 is 0.0668

ii)

Proportion of the values that will be exceeds -2.0

P( Z > -2 ) = 1 - P( Z <= -2 )

Using Excel, P( z <= -2) = NORMSDIST(-2) = 0.02275

P( Z > -2 ) = 1 - 0.02275 = 0.9773

Proportion of the values that will be exceeds -2.0 is 0.9773 .

c)

We have to find P( -1.23 < Z < 2.85 )

P( -1.23 < Z < 2.85 ) = P( Z < 2.85 ) - P( Z < -1.23 )

Using Excel,

P( -1.23 < Z < 2.85 ) = NORMSDIST(2.85) - NORMSDIST(-1.23) = 0.8885

Proportion of the values that will between -1.23 and 2.85 is 0.8885

d )

i ) Proportion of values that will be exceeds 5

P( Z > 5 ) = 1 - P( Z < 5 )

Using Excel, P( z <= 5) = NORMSDIST(5) = 0.999999713 $\approx$ 1

P( Z > 5 ) = 1 - 1 = 0

Proportion of values that will be exceeds 5 is zero.

ii) Proportion of values that will be exceeds -5

P( Z > -5 ) = 1 - P( Z < -5 )

Using Excel, P( z <= -5) = NORMSDIST-(5) =0.0000002 $\approx$ 0

P( Z > -5 ) = 1 - 0 =1

Proportion of values that will be exceeds -5 is 1.

e )

We have to find P( |Z| < 2.50 )

P( |Z| < 2.50 ) = P( -2.50 < Z < 2.50 )

P( -2.50 < Z < 2.50 ) = P( Z < 2.50 ) - P( Z < - 2.50 )

Using Excel,

P( -2.50 < Z < 2.50) = NORMSDIST(2.50) - NORMSDIST(-2.50) = 0.9876

Proportion of selected values z will be satisfy |z| < 2.50 is 0.9876