Question

Assume a normal distribution and find the following probabilities. (Round the values of z to 2 decimal places. Round your answers to 4 decimal places.) (a) P(x < 23 | μ = 26 and σ = 4) enter the probability of fewer than 23 outcomes if the mean is 26 and the standard deviation is 4 (b) P(x ≥ 42 | μ = 30 and σ = 7) enter the probability of 42 or more outcomes if the mean is 30 and the standard deviation is 7 (c) P(x > 23 | μ = 30 and σ = 6) enter the probability of more than 23 outcomes if the mean is 30 and the standard deviation is 6 (d) P(22 < x < 25 | μ = 24 and σ = 4) enter the probability of more than 22 and fewer than 25 outcomes if the mean is 24 and the standard deviation is 4 (e) P(x ≥ 95 | μ = 80 and σ = 2.80)

Answer 1

Here normal distribution is assumed.

Hence the z score formula is given by,

a) P(x < 23 | μ = 26 and σ = 4)

p ( x < 23 )

----------( converted in to z score )

= p ( z < -0.75 )

= 0.2266 ------------( We get this probability using excel formula " =norm.s.dist(-0.75,1)" )

b) P(x ≥ 42 | μ = 30 and σ = 7)

P(x ≥ 42 )

-------- ( converted in to z score )

= p ( z  ≥ 1.71 )

= 1 - p ( z < 1.71 ) --------------( using distribution function property )

= 1 - 0.9564   ------------( We get this probability using excel formula " =norm.s.dist(1.71,1)" )

= 0.0436

c) P(x > 23 | μ = 30 and σ = 6)

p ( x > 23 )

   -------- ( converted in to z score )

= p ( z >  -1.17 )  

= 1 - p( z -1.17 )   --------------( using distribution function property )

= 1 - 0.1210    ------------( We get this probability using excel formula " =norm.s.dist(-1.17,1)" )

= 0.8790

d) P(22 < x < 25 | μ = 24 and σ = 4)

P(22 < x < 25 )

-------- ( converted in to z score )

= p ( -0.5 < z < 0.25 )

= p ( z < 0.25 ) - p ( z < - 0.5 ) --------------( using distribution function property )

= 0.5987 - 0.3085  ------------( We get these probabilities using excel formula " =norm.s.dist(0.25,1)" n "=norm.s.dist(-0.5,1)" )

= 0.2902

e) We need to find, P(x ≥ 95 | μ = 80 and σ = 2.80)

P(x ≥ 95)

-------- ( converted in to z score )

= p ( z  ≥ 5.36 )

= 1 - p ( z < 5.36 ) --------------( using distribution function property )

= 1 -1 ------------( We get this probability using excel formula " =norm.s.dist(5.36,1)" )

= 0.0000