Question

1. If Z is a standard normal random variable, find c such that P(−c ≤ Z ≤ c) = 0.82. [Answer to 2 decimal places]

2. Weakly earnings on a certain import venture are approximately normally distributed with a known mean of $353 and unknown standard deviation. If the proportion of earnings over $386 is 25%, find the standard deviation. Answer only up to two digits after decimal.

3. X is a normal random variable with mean μ and standard deviation σ. Then P( μ− 1.7 σ ≤ X ≤ μ+ 2.8 σ) =? Answer to 4 decimal places.

Answer 1

Solution,

1) Using standard normal table,

P( -c Z c) = 0.82

= P(Z c) - P(Z -c ) = 0.82

= 2P(Z c) - 1 = 0.82

= 2P(Z c) = 1 + 0.82

= P(Z c) = 1.82 / 2

= P(Z c) = 0.91

= P(Z 1.34) = 0.91

= c ± 1.34

2) Given that,

mean = = 353

x = 386

Using standard normal table,

P(Z > z) = 25%

= 1 - P(Z < z) = 0.25

= P(Z < z) = 1 - 0.25

= P(Z < z ) = 0.75

= P(Z < 0.6745 ) = 0.75

z = 0.6745

Using z-score formula,

x = z * +

386 = 0.6745 * + 353

= 386 - 353 / 0.6745

= 48.93

3) P( -1.7 < Z < 2.8)

= P( Z < 2.8) - P( Z < -1.7)

= 0.9974 - 0.0446

= 0.9528