Math Problem Statement

Z is the standard normal variable. Find the indicated probability. (Round your answer to four decimal places.)

P(0 ≤ Z ≤ 0.6)

Solution

To find the probability P(0Z0.6)P(0 \leq Z \leq 0.6) for a standard normal variable ZZ, follow these steps:

Step 1: Look up cumulative probabilities in the Z-table.

The Z-table provides the cumulative probability P(Zz)P(Z \leq z) from the leftmost tail to a given Z-value.

  • P(Z ≤ 0): Since the Z-table is symmetric, P(Z0)=0.5P(Z \leq 0) = 0.5.
  • P(Z ≤ 0.6): From the Z-table, the value for Z=0.6Z = 0.6 is 0.7257.

Step 2: Calculate P(0Z0.6)P(0 \leq Z \leq 0.6).

Since P(0Z0.6)P(0 \leq Z \leq 0.6) is the area between Z=0Z = 0 and Z=0.6Z = 0.6, we subtract:

P(0Z0.6)=P(Z0.6)P(Z0)P(0 \leq Z \leq 0.6) = P(Z \leq 0.6) - P(Z \leq 0)

P(0Z0.6)=0.72570.5000=0.2257P(0 \leq Z \leq 0.6) = 0.7257 - 0.5000 = 0.2257

Final Answer:

P(0Z0.6)=0.2257P(0 \leq Z \leq 0.6) = 0.2257

Would you like further explanation on using Z-tables, or do you have additional questions?


Here are 5 related questions:

  1. What is the probability for P(Z1.2)P(Z \geq 1.2)?
  2. How do you compute P(1Z1)P(-1 \leq Z \leq 1)?
  3. What is the total area under the standard normal curve?
  4. How can we use the Z-table to find P(Z0.8)P(Z \leq -0.8)?
  5. How does symmetry help when computing probabilities for Z-values around 0?

Tip: Always subtract cumulative probabilities carefully to avoid calculation errors when finding areas between two Z-values.

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Math Problem Analysis

Mathematical Concepts

Probability
Statistics
Standard Normal Distribution
Z-Score

Formulas

P(a ≤ Z ≤ b) = P(Z ≤ b) - P(Z ≤ a)
Standard Normal Distribution formula

Theorems

Properties of the Standard Normal Distribution
Symmetry of the Normal Curve

Suitable Grade Level

Grades 10-12