Math Problem Statement

In a Standard Normal Distribution, find the probabilities for specific Z-values and compute various probabilities for a standard normal distribution, rounding to the required decimal places.

Solution

Let's go through each part of Questions 5 and 6, using the properties of the standard normal distribution (Z-distribution) and a Z-table to find the probabilities.

Question 5

Given: Standard Normal Distribution, rounded to 3 decimal places.

  1. (a) P(Z>1.96)P(Z > 1.96)

    • Since ZZ is greater than a specific value, we need the area to the right of Z=1.96Z = 1.96.
    • The Z-table shows that P(Z<1.96)=0.975P(Z < 1.96) = 0.975.
    • Thus, P(Z>1.96)=10.975=0.025P(Z > 1.96) = 1 - 0.975 = 0.025.
  2. (b) P(Z<0.8)P(Z < 0.8)

    • We need the area to the left of Z=0.8Z = 0.8.
    • The Z-table gives P(Z<0.8)=0.7881P(Z < 0.8) = 0.7881.
  3. (c) P(0.6<Z<1.2)P(-0.6 < Z < 1.2)

    • We calculate P(Z<1.2)P(Z < 1.2) and P(Z<0.6)P(Z < -0.6) and subtract them.
    • From the Z-table:
      • P(Z<1.2)=0.8849P(Z < 1.2) = 0.8849
      • P(Z<0.6)=0.2743P(Z < -0.6) = 0.2743
    • Therefore, P(0.6<Z<1.2)=0.88490.2743=0.6106P(-0.6 < Z < 1.2) = 0.8849 - 0.2743 = 0.6106.

Question 6

Given: Standard Normal Distribution, rounded to at least 4 decimal places.

  1. (a) P(Z>0.23)P(Z > -0.23)

    • We find P(Z<0.23)P(Z < -0.23) and subtract from 1.
    • The Z-table shows P(Z<0.23)=0.4090P(Z < -0.23) = 0.4090.
    • So, P(Z>0.23)=10.4090=0.5910P(Z > -0.23) = 1 - 0.4090 = 0.5910.
  2. (b) P(2.7<Z<0.45)P(-2.7 < Z < 0.45)

    • Find P(Z<0.45)P(Z < 0.45) and P(Z<2.7)P(Z < -2.7), then subtract.
    • From the Z-table:
      • P(Z<0.45)=0.6736P(Z < 0.45) = 0.6736
      • P(Z<2.7)=0.0035P(Z < -2.7) = 0.0035
    • Thus, P(2.7<Z<0.45)=0.67360.0035=0.6701P(-2.7 < Z < 0.45) = 0.6736 - 0.0035 = 0.6701.
  3. (c) P(Z<1.57)P(Z < 1.57)

    • From the Z-table, P(Z<1.57)=0.9418P(Z < 1.57) = 0.9418.
  4. (d) P(Z2.89)P(Z \geq 2.89)

    • We find P(Z<2.89)P(Z < 2.89) and subtract from 1.
    • From the Z-table, P(Z<2.89)=0.9981P(Z < 2.89) = 0.9981.
    • So, P(Z2.89)=10.9981=0.0019P(Z \geq 2.89) = 1 - 0.9981 = 0.0019.

Summary of Answers

  • Question 5:

    • (a) P(Z>1.96)=0.025P(Z > 1.96) = 0.025
    • (b) P(Z<0.8)=0.788P(Z < 0.8) = 0.788
    • (c) P(0.6<Z<1.2)=0.611P(-0.6 < Z < 1.2) = 0.611
  • Question 6:

    • (a) P(Z>0.23)=0.5910P(Z > -0.23) = 0.5910
    • (b) P(2.7<Z<0.45)=0.6701P(-2.7 < Z < 0.45) = 0.6701
    • (c) P(Z<1.57)=0.9418P(Z < 1.57) = 0.9418
    • (d) P(Z2.89)=0.0019P(Z \geq 2.89) = 0.0019

Would you like a deeper explanation of any step, or do you have any questions?

Related Questions to Explore

  1. How would the probabilities change if we used a different Z-score, like Z>2.5Z > 2.5?
  2. What are the steps to calculate probabilities without a Z-table?
  3. How do you interpret a Z-score in real-world contexts?
  4. What is the probability of a Z-score falling between two specific values (e.g., P(1<Z<2)P(1 < Z < 2))?
  5. Can we find probabilities for non-standard normal distributions?

Tip

When using a Z-table, remember that it typically gives the area to the left of a given Z-score, so you may need to adjust your calculation for right-tail probabilities.

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Math Problem Analysis

Mathematical Concepts

Probability
Statistics
Standard Normal Distribution
Z-scores

Formulas

P(Z > z) = 1 - P(Z < z)
P(a < Z < b) = P(Z < b) - P(Z < a)

Theorems

Standard Normal Distribution (Z-distribution)

Suitable Grade Level

Grades 11-12, College Statistics