Question

Calculate the required probabilities for the normal distributions with the parameters specified in parts a through e. a. u = 5,0 = 4; calculate P(0 < x < 6). P(0<x< 6) = (Round to four decimal places as needed.) b. = 5,0 = 5; calculate P(0 < x < 6). PlO< x < 6) = (Round to four decimal places as needed.) C. M = 4,0 = 4; calculate P(0 < x < 6). P(0 < x < 6) = (Round to four decimal places as needed.) d. u = 6,0 = 2; calculate P(x > 1). P(x > 1) = .(Round to four decimal places as needed.) e. U = 0, 0 = 2; calculate P(x > 1). P(x > 1) = (Round to four decimal places as needed.)

Answer 1

Solution :

a )Given that,

mean = $\mu$ = 5

standard deviation = $\sigma$ = 4

P (0 < x < 6 )

P ( 0 - 5 / 4) < ( x -  $\mu$/ $\sigma$ ) < ( 6 - 5 /4)

P ( -5 / 4 < z < 1 / 4 )

P (-1.25 < z < 0.25)

P ( z < 0.25 ) - P ( z < -1.25)

Using z table

= 0.5987 - 0.1056

= 0.4931

Probability = 0.4931

b )Given that,

mean = $\mu$ = 5

standard deviation = $\sigma$ = 5

P (0 < x < 6 )

P ( 0 - 5 / 5) < ( x -  $\mu$/ $\sigma$ ) < ( 6 - 5 / 5)

P ( - 5 / 5 < z < 1 / 5 )

P (-1 < z < 0.2)

P ( z < 0.2 ) - P ( z < -1)

Using z table

= 0.5793 - 0.1587

= 0.4206

Probability = 0.4206

c ) Given that,

mean = $\mu$ = 4

standard deviation = $\sigma$ = 4

P (0 < x < 6 )

P ( 0 - 4 / 4) < ( x -  $\mu$/ $\sigma$ ) < ( 6 - 4 / 4)

P ( - 4 / 4 < z < 2 / 4 )

P (-1 < z < 0.5)

P ( z < 0.5 ) - P ( z < -1)

= 0.6915 - 0.1587

= 0.5328

Probability = 0.5328

d ) Given that,

mean = $\mu$ = 6

standard deviation = $\sigma$ = 2

P (x > 1 )

= 1 - P (x < 1 )

= 1 - P ( x -  $\mu$/ $\sigma$ ) < ( 1 -6 / 2)

= 1 - P ( z <- 5 / 2 )

= 1 - P ( z < -2.5)

Using z table

= 1 - 0.0062

= 0.9938

Probability = 0.9938

e ) Given that,

mean = $\mu$ = 0

standard deviation = $\sigma$ = 2

P (x > 1 )

= 1 - P (x < 1 )

= 1 - P ( x -  $\mu$/ $\sigma$ ) < ( 1 - 0 / 2)

= 1 - P ( z < 1 / 2 )

= 1 - P ( z < 0.50)

Using z table

= 1 - 0.6915

= 0.3085

Probability = 0.3085