Question

Q2. Given a normal distribution with μ = 30 and σ = 6,
find
1- the normal curve area to the right of x = 17 [Hint: P(X>17)]

2-the normal curve area to the left of x = 22 [Hint: P(X<22)]

3-the normal curve area between x = 32 and x = 41[Hint: P(32<X<41)];

4-the value of x that has 80% of the normal curve area to the left [Hint: P(X<k)=0.8];

Answer 1

= 30

= 6

P(X < A) = P(Z < (A - )/)

1) P(X > 17) = 1 - P(X < 17)

= 1 - P(Z < (17 - 30)/6)

= 1 - P(Z < -2.17)

= 1 - 0.0150

= 0.9850

2) P(X < 22) = P(Z < (22 - 30)/6)

= P(Z < -1.33)

= 0.0918

3) P(32 < X < 41) = P(X < 41) - P(X < 32)

= P(Z < (41 - 30)/6) - P(Z < (32 - 30)/6)

= P(Z < 1.83) - P(Z < 0.33)

= 0.9664 - 0.6293

= 0.3371

4) P(X < k) = 0.8

P(Z < (k - 30)/6) = 0.8

Take value of z corresponding to 0.8 from standard normal distribution table

(k - 30)/6 = 0.84

k = 35.04