Question

Calculate the area under the standard normal curve to the left of these values. (Round your answers to four decimal places.)

(a)

z = 1.6

(b)

z = 1.89

(c)

z = 0.60

(d)

z = 4.18

Find the following probabilities for the standard normal random variable z. (Round your answers to four decimal places.)

(a)    

P(−1.46 < z < 0.66) =

(b)    

P(0.52 < z < 1.76) =

(c)    

P(−1.52 < z < −0.49) =

(d)    

P(z > 1.32) =

(e)    

P(z < −4.36) =

(a)

Find a z₀ such that

P(z > z₀) = 0.0301.

(Round your answer to two decimal places.)

z₀ =

(b)

Find a z₀ such that

P(z < z₀) = 0.9192.

(Round your answer to two decimal places.)

z₀ =

Answer 1

We will find the solution using R-studio

(a)

z = 1.6

> pnorm(1.6)
[1] 0.9452007

so P(Z< 1.6) = 0.9452

(b)

z = 1.89

> pnorm(1.89)
[1] 0.970621

so P(Z< 1.89) = 0.9706

(c)

z = 0.60

> pnorm(0.60)
[1] 0.7257469

so P(Z< 0.60) =0.7258

(d)

z = 4.1

> pnorm(0.41)
[1] 0.659097

so P(Z< 0.41) = 0.6591

Find the following probabilities for the standard normal random variable z. (Round your answers to four decimal places.)

(a)    

P(−1.46 < z < 0.66) =0.6732

> round(pnorm(0.66)-pnorm(-1.46),4)
[1] 0.6732

(b)    

P(0.52 < z < 1.76) = 0.2623

> round(pnorm(1.76)-pnorm(0.52),4)
[1] 0.2623

(c)    

P(−1.52 < z < −0.49) =0.2478

> round(pnorm(-0.49)-pnorm(-1.52),4)
[1] 0.2478

(d)    

P(z > 1.32) = 0.0934

> round(1-pnorm(1.32),4)
[1] 0.0934

(e)    

P(z < −4.36) = 0

> round(pnorm(-4.36),4)
[1] 0

(a)

Find a z₀ such that

P(z > z₀) = 0.0301.

> round(qnorm(0.0301),2)
[1] -1.88

z₀ = -1.88

(b)

Find a z₀ such that

P(z < z₀) = 0.9192.

> round(qnorm(0.9192),2)
[1] 1.4

z₀ =1.4