Math Problem Statement

When working on a normal distribution other than the standard normal distribution you need to either first convert to the z-scores and then find probabilities using the standard normal distribution(N(0,1)) OR when using technology you can input the mean and standard deviation.

P (a < X < b) = normalcdf(a,b,)=normalcdf(staring data value, ending data value, mean, standard deviation) to find the probability given the interval of data values.

This is equal to P(<Z<) = normalcdf(,)=normalcdf(smaller z-score, larger z-score)

For finding the data value given the area you can use invNorm. Again you either need to convert the z-score back to the data value OR tell the calculator the mean and standard deviation.

invNorm(area to the left, mean, standard deviation) = data value =X

invNorm(area to the left) = Z = Z-score. Then X =

For the distribution N(10,2) match the probabilities below:

Group of answer choices P(X < a) = 0.1

P(X > a) = 0.1

Solution

To solve these probability questions for the normal distribution N(10,2)N(10, 2), we need to use the concepts of normalcdf and invNorm as described. Let’s break this down:

Given Information:

  • The normal distribution is N(10,2)N(10, 2), where:
    • μ=10\mu = 10 (mean)
    • σ=2\sigma = 2 (standard deviation)

We are tasked with finding the value of aa that corresponds to:

  1. P(X<a)=0.1P(X < a) = 0.1
  2. P(X>a)=0.1P(X > a) = 0.1

Steps for Solution:

1. P(X<a)=0.1P(X < a) = 0.1

This means we are looking for the value of aa such that the area under the normal curve to the left of aa is 0.1. We can use the invNorm function to find this.

  • For the normal distribution N(10,2)N(10, 2), we input: a=invNorm(0.1,10,2)a = \text{invNorm}(0.1, 10, 2)

The invNorm function gives us the aa-value corresponding to a cumulative probability (area) of 0.1.

2. P(X>a)=0.1P(X > a) = 0.1

This means we are looking for the value of aa such that the area under the normal curve to the right of aa is 0.1. To find this, we first convert it to a cumulative probability (area to the left of aa).

  • Since P(X>a)=0.1P(X > a) = 0.1, the cumulative probability to the left of aa is 10.1=0.91 - 0.1 = 0.9.
  • So, we use: a=invNorm(0.9,10,2)a = \text{invNorm}(0.9, 10, 2)

Solutions:

  1. P(X<a)=0.1P(X < a) = 0.1: Solve a=invNorm(0.1,10,2)a = \text{invNorm}(0.1, 10, 2).
  2. P(X>a)=0.1P(X > a) = 0.1: Solve a=invNorm(0.9,10,2)a = \text{invNorm}(0.9, 10, 2).

Do you have a calculator, or would you like me to compute these values for you?

Relative Questions:

  1. What is the relationship between the z-score and the normal distribution parameters?
  2. How would you calculate probabilities for a range P(a<X<b)P(a < X < b) in this distribution?
  3. Can you explain the difference between cumulative probability and probability density?
  4. What happens to the shape of the normal distribution as standard deviation increases?
  5. How would you find the z-score for a particular data value?

Tip:

Always make sure your calculator's mode is set to the correct distribution (normal, binomial, etc.) when performing statistical calculations!

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Probability
Z-scores

Formulas

P(X < a) = invNorm(area, mean, standard deviation)
P(X > a) = 1 - invNorm(area, mean, standard deviation)
P(a < X < b) = normalcdf(start, end, mean, standard deviation)

Theorems

Normal Distribution Theorem
Z-Score Calculation

Suitable Grade Level

Grades 10-12