Answer :
To solve these problems, we will use properties of the normal distribution. Given that X is normally distributed with a mean [tex]\mu = 12[/tex] and a standard deviation [tex]\sigma = 4[/tex], we can standardize any value 'x' using the formula [tex]Z = \frac{x - \mu}{\sigma}[/tex] to find the corresponding Z-score and then use Z-tables. We assume the provided Z values correspond to probabilities.
Probability that X ≥ 20:
First, find the Z-score for [tex]X = 20[/tex]:
[tex]Z = \frac{20 - 12}{4} = \frac{8}{4} = 2[/tex]Now, look up the probability for Z = 2. Typically, Z-tables give the probability that a value is less than a given Z-score (i.e., P(Z ≤ 2)). Assuming you have the Z-table, the probability [tex]P(Z \leq 2)[/tex] is approximately 0.9772 (this is a standard value, derived from cumulative distribution tables).
Therefore, [tex]P(X \geq 20) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228[/tex].
Probability that X ≤ 20:
From the earlier calculation, [tex]P(X \leq 20) = P(Z \leq 2) = 0.9772[/tex].
Probability that 0 ≤ X ≤ 12:
First, find the Z-score for [tex]X = 0[/tex]:
[tex]Z = \frac{0 - 12}{4} = \frac{-12}{4} = -3[/tex]
(This extremely low Z-score suggests using direct table values isn't straightforward, as the probabilities are very small)The Z-score for [tex]X = 12[/tex] is:
[tex]Z = \frac{12 - 12}{4} = 0[/tex]
From standard tables, [tex]P(Z \leq 0) = 0.5[/tex].To find [tex]P(0 \leq X \leq 12)[/tex], calculate:
[tex]P(0 \leq X \leq 12) = P(Z \leq 0) - P(Z \leq -3)[/tex]
As [tex]P(Z \leq -3)[/tex] is very close to 0 (commonly [tex]\approx 0.0013[/tex]), it is often negligible in comparative analysis with [tex]P(Z \leq 0)[/tex]. For a precise number consult specific Z-tables. Thus, [tex]P(0 \leq X \leq 12) \approx 0.5[/tex].
