Answer:
a) P(x > 120) = 0.03288
b) P(100 < x < 110) = 0.46777
c) value of a for which P(x < a) = 0.25
a = 99.91
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 105.3
Standard deviation = σ = 8.0
a) P(x > 120)
To work this, we first need To convert 120 go standardized/normalized/z-scores.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (120 - 105.3)/8.0 = 1.84
To determine P(x > 120) = P(z > 1.84)
We use the tables for these probabilities now.
P(x > 120) = P(z > 1.84) = 1 - P(z ≤ 1.84) = 1 - 0.96712 = 0.03288
b) P(100 < x < 110)
We normalize the two value in the inequality
For 100,
z = (x - μ)/σ = (100 - 105.3)/8.0 = -0.66
For 110,
z = (x - μ)/σ = (110 - 105.3)/8.0 = 0.59
To determine P(100 < x < 110) = P(-0.66 < z < 0.59)
We use the tables for these probabilities now.
P(100 < x < 110) = P(-0.66 < z < 0.59) = P(z < 0.59) - P(z < -0.66) = 0.72240 - 0.25463 = 0.46777
c) value of a for which P(x < a) = .25
Let the z-score of a be z'
z' = (x - μ)/σ = (a - 105.3)/8.0
P(x < a) = P(z < z') = 0.25
Using the table, z' = - 0.674
-0.674 = (a - 105.3)/8
a - 105.3 = -5.392
a = 105.3 - 5.392 = 99.91
Hope this Helps!!!
