Find z if the standard normal curve area between -z and z is 0.4530?

Hi,

1) Find z if the standard normal curve area between -z and z is 0.4530.

Y1 = normalcdf(-x,x,0,1) - .4530 and x = .60226

a. ± 0.118

b. ± 0.398

c. ± 0.602 <==ANSWER

d. ± 0.752

2) If a random variable has the normal distribution with μ = 104.3 and σ = 5.7, find the probability that the value will be greater than 112.3

normalcdf(112.3, 500, 104.3, 5.7) = .08023

a. 0.08 <==ANSWER

b. 0.09

c. 0.10

d. 0.11

3) Determine a 90% confidence interval for μ if σ = 5, \bar{x} = 70, and n = 82. Answer interval rounded to two decimal places.

70 ± 1.645 (5/√82) or 69.09 to 70.91 <==ANSWER

I hope that helps!! :-)

I will do the first one,1) If the area is 0.4530, then the area > z(assuming z> 0 )will be ( 1 - .4530)/2 = 0.2735

Hence the area < z is 1 - 0.2735 = 0.7265. From tables we get z = 0.602 so c).

1) C

2) A

3) (69.09,70.91)

Work

____________________________________

1) normalcdf(-.602,.602) = .4530

2)

P(x>112.3)

z = (x - μ) / σ

z = (112.3 - 104.3) / 5.7 = 1.40351

P(z>1.40351) = normalcdf(1.40351,100) = .08

3) CI = x +/- z(σ/√n)

.90CI = 70 +/- 1.645(5/√82)

.90CI = (69.09,70.91)

Z 104.3