Question

How do I do these computations without a normal distribution chart and just a TI84 calculator?

6.7 Given a standard normal distribution, find the value of k such that

(a) P(Z>k)=0 .2946;

(b) P(Z<k)=0 .0427;

(c) P(−0.93 <Z<k)=0 .7235.

6.8 Given a normal distribution with μ = 30 and σ = 6, find

(a) the normal curve area to the right of x = 17;

(b) the normal curve area to the left of x = 22;

(c) the normal curve area between x = 32 and x = 41;

(d) the value of x that has 80% of the normal curve area to the left;

(e) the two values of x that contain the middle 75% of the normal curve area.

Answer 1

Solution:-

6.8)

μ = 30 and σ = 6

a) The normal curve area to the right of x = 17 is 0.9849.

x = 17

By applying normal distribution:-

z = - 2.167

P(z > -2.167) = 0.9849

b) The normal curve area to the left of x = 22 is 0.0913.

x = 22

By applying normal distribution:-

z = - 1.333

P(z < -1.33) = 0.0913

c) The normal curve area between x = 32 and x = 41 is 0.3361.

x₁ = 32

x₂ = 41

By applying normal distribution:-

z₁ = 0.3333

z₂ = 1.8333

P( 0.333 < z < 1.833) = P(z > 0.333) - P(z > 1.8333)  

P( 0.333 < z < 1.833) = 0.3695 - 0.0334

P( 0.333 < z < 1.833) = 0.3361

d) The value of x that has 80% of the normal curve area to the left is 35.052.

p-value for the bottom 80% = 0.80

z-score for the p-value = 0.842

By applying normal distribution:-

x = 35.052

e) The two values of x that contain the middle 75% of the normal curve area is 23.1 and 36.9

p-value for the middle 75% = 0.125 and 0.875

z-score for the p-value = + 1.15

By applying normal distribution:-

x₁ = 23.1

x₂ = 36.9