Answer :
Let's break down the question step by step. We are given that [tex]X[/tex] is a normally distributed random variable with mean [tex]\mu = 12[/tex] and standard deviation [tex]\sigma = 4[/tex]. This is denoted as [tex]X \sim N(12, 4^2)[/tex].
Part a
We need to find the probabilities for different scenarios.
(i)
[tex]P(X \geq 20)[/tex]To find the probability of [tex]X[/tex] being greater than or equal to 20, we first convert this to the standard normal variable [tex]Z[/tex] using:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
For [tex]X = 20[/tex]:
[tex]Z = \frac{20 - 12}{4} = 2[/tex]
Thus, [tex]P(X \geq 20) = P(Z \geq 2)[/tex].
Using a standard normal distribution table or calculator, [tex]P(Z \geq 2)[/tex] is approximately 0.0228.
(ii)
[tex]P(X \leq 20)[/tex]This is simply the complement of the previous probability:
[tex]P(X \leq 20) = 1 - P(X > 20) = 1 - 0.0228 = 0.9772[/tex]
(iii)
[tex]P(0 \leq X \leq 12)[/tex]First, we find [tex]P(X \leq 12)[/tex]:
[tex]Z = \frac{12 - 12}{4} = 0[/tex]
From standard normal distribution table, [tex]P(Z \leq 0) = 0.5[/tex].
Then, [tex]P(X \geq 0)[/tex]:
[tex]Z = \frac{0 - 12}{4} = -3[/tex]
Referencing a standard normal distribution table, [tex]P(Z \leq -3)[/tex] is approximately 0.0013.
Hence, [tex]P(X \geq 0) = 1 - 0.0013 = 0.9987[/tex].
Thus, [tex]P(0 \leq X \leq 12) = P(X \leq 12) - P(X < 0) = 0.5 - (1 - 0.9987) = 0.5 - 0.0013 = 0.4987[/tex].
Part b
We need to find [tex]x'[/tex] such that [tex]P(X > x') = 0.24[/tex].
This means [tex]P(X \leq x') = 1 - 0.24 = 0.76[/tex].
Looking up [tex]0.76[/tex] in a standard normal distribution table or using inverse normal calculations gives approximately [tex]Z = 0.706[/tex].
Now, convert [tex]Z[/tex] back to [tex]X[/tex]:
[tex]x' = \mu + Z \cdot \sigma[/tex]
[tex]x' = 12 + 0.706 \times 4 \approx 14.824[/tex]
Thus, [tex]x' \approx 14.82[/tex].
