Math Problem Statement

assume a standard normal distribution for these calculations. Go through calculations in detail

Solution

Great! We'll assume a standard normal distribution for these calculations, which means the mean (μ\mu) is 0 and the standard deviation (σ\sigma) is 1. The values given are rates, and we'll treat them as z-scores directly. Here's how we can solve each part:

1. Find the probability that a randomly selected rate is higher than 4.75%.

Since we're using a standard normal distribution:

  • The z-score corresponding to 4.75% is directly 4.75.
  • We want to find P(Z>4.75)P(Z > 4.75).

Using standard normal distribution tables or a calculator:

  • The probability P(Z4.75)P(Z \leq 4.75) is almost 1 (since 4.75 is far into the tail).
  • Therefore, P(Z>4.75)P(Z > 4.75) is almost 0.

P(Z>4.75)0P(Z > 4.75) \approx 0

2. Find the probability that a randomly selected rate is at most 4.52%.

Again, using the standard normal distribution:

  • The z-score corresponding to 4.52% is directly 4.52.
  • We want to find P(Z4.52)P(Z \leq 4.52).

From the standard normal distribution table:

  • P(Z4.52)1P(Z \leq 4.52) \approx 1 (since 4.52 is very high).

P(Z4.52)1P(Z \leq 4.52) \approx 1

3. Find the probability that a randomly selected rate is between 4.64% and 5.26%.

Here, we calculate two probabilities:

  • P(Z5.26)P(Z \leq 5.26) and P(Z4.64)P(Z \leq 4.64).
  • The probability we want is P(4.64Z5.26)=P(Z5.26)P(Z4.64)P(4.64 \leq Z \leq 5.26) = P(Z \leq 5.26) - P(Z \leq 4.64).

Using standard normal distribution tables or a calculator:

  • P(Z5.26)1P(Z \leq 5.26) \approx 1 (since 5.26 is very high).
  • P(Z4.64)1P(Z \leq 4.64) \approx 1 (since 4.64 is also very high).

P(4.64Z5.26)=11=0P(4.64 \leq Z \leq 5.26) = 1 - 1 = 0

4. Find the rate corresponding to a standard score (z-score) of 0.85.

Given:

  • z=0.85z = 0.85
  • The formula to find the corresponding rate (X) is: X=μ+z×σX = \mu + z \times \sigma

Since μ=0\mu = 0 and σ=1\sigma = 1: X=0+0.85×1=0.85X = 0 + 0.85 \times 1 = 0.85

So, the rate corresponding to a z-score of 0.85 is 0.85 or 85%.

Summary:

  1. Probability of rate higher than 4.75%: approximately 0.
  2. Probability of rate at most 4.52%: approximately 1.
  3. Probability of rate between 4.64% and 5.26%: 0.
  4. Rate corresponding to a z-score of 0.85: 0.85 or 85%.

Would you like further clarification on any of these points?

Follow-up Questions:

  1. How would the probabilities change if the rates were closer to the mean?
  2. What does a z-score of 0.85 signify in a standard normal distribution?
  3. How would you find probabilities if the mean and standard deviation were different?
  4. What is the significance of tail probabilities in a normal distribution?
  5. How do these results apply to real-world scenarios?

Tip:

When dealing with probabilities in a normal distribution, remember that extreme z-scores (far from 0) correspond to events that are very rare.

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Standard Normal Distribution
Z-score

Formulas

Z-score formula: \( Z = \frac{X - \mu}{\sigma} \)

Theorems

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Suitable Grade Level

Advanced High School / College Level