Questions: d. P(z ≥ -2.5) e. P(-3 < z ≤ 0)

d. P(z ≥ -2.5)
e. P(-3 < z ≤ 0)
Transcript text: d. $P(z \geq-2.5)$ e. $P(-3
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate \( P(z \geq -2.5) \)

To find the probability \( P(z \geq -2.5) \), we can express it using the cumulative distribution function \( \Phi \):

\[ P(z \geq -2.5) = 1 - P(z < -2.5) = 1 - \Phi(-2.5) \]

Using the properties of the standard normal distribution, we have:

\[ P(z \geq -2.5) = \Phi(\infty) - \Phi(-2.5) = 0.9938 \]

Thus, the result for part d is:

\[ \boxed{P(z \geq -2.5) = 0.9938} \]

Step 2: Calculate \( P(-3 < z \leq 0) \)

To find the probability \( P(-3 < z \leq 0) \), we can express it as:

\[ P(-3 < z \leq 0) = P(z \leq 0) - P(z \leq -3) = \Phi(0) - \Phi(-3) \]

Using the cumulative distribution function values, we have:

\[ P(-3 < z \leq 0) = \Phi(0) - \Phi(-3) = 0.4987 \]

Thus, the result for part e is:

\[ \boxed{P(-3 < z \leq 0) = 0.4987} \]

Final Answer

The answers to the questions are:

  • For part d: \( \boxed{P(z \geq -2.5) = 0.9938} \)
  • For part e: \( \boxed{P(-3 < z \leq 0) = 0.4987} \)
Was this solution helpful?
failed
Unhelpful
failed
Helpful