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X is normally distributed with a mean of 12 and a standard deviation of 4.

Find out probabilities that:

1) X ≥ 20

2) X ≤ 20

3) 0 ≤ X ≤ 12

(Given: Z(0.2) = 0.4772, Z(0.1) = 0.3413, Z(0.3) = 0.49865)

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.

  1. 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].

  2. Probability that X ≤ 20:

    From the earlier calculation, [tex]P(X \leq 20) = P(Z \leq 2) = 0.9772[/tex].

  3. 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].

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