Math Problem Statement

The document includes two exercises about normal distribution and confidence intervals. Exercise 1 concerns the first words spoken by children following a normal distribution with a mean of 12 months and a standard deviation of 2.5 months. Exercise 2 involves calculating confidence intervals for the mean of a concrete material's compression force with sample data provided.

Solution

The document you uploaded contains two exercises related to statistical concepts, specifically dealing with the normal distribution and confidence intervals. Let's break down the content and work on the solutions for each question.

Exercise 1: Normal Distribution

It describes a situation where the age at which a child speaks their first words follows a normal distribution with:

  • Mean (μ) = 12 months
  • Standard deviation (σ) = 2.5 months
  1. Identify the population, variable, its type, and parameters:

    • Population: Children
    • Variable: Age at which children speak their first words
    • Type of variable: Continuous variable
    • Parameters: Mean (μ) = 12 months, Standard deviation (σ) = 2.5 months
  2. What is the proportion of children for whom the first words appear before 9 months? To find this, we will use the Z-score formula for normal distribution: Z=XμσZ = \frac{X - \mu}{\sigma} where X=9X = 9 months, μ=12\mu = 12 months, and σ=2.5\sigma = 2.5 months.

  3. Determine the age above which 2% of children speak their first words. We can use the Z-table to find the Z-value corresponding to the cumulative area of 0.02 (for 2% in the tail). Then, we can solve for the corresponding age.


Exercise 2: Confidence Interval of the Mean

This question involves calculating confidence intervals for the mean compression force of a type of concrete, modeled by a normal distribution.

  • The variance is given as σ2=1000\sigma^2 = 1000, and the sample size is n=12n = 12.
  • The sample mean is xˉ=3250\bar{x} = 3250 psi.
  1. Construct a 95% confidence interval for the mean (μ). The formula for a confidence interval is: μ[xˉzα/2×σn,xˉ+zα/2×σn]\mu \in \left[ \bar{x} - z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}, \bar{x} + z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} \right] where zα/2z_{\alpha/2} corresponds to the Z-score for a 95% confidence level.

  2. Construct a 99% confidence interval for the mean (μ). Similar to the 95% interval but with the Z-value corresponding to 99% confidence.

  3. For the same sample, find a 30 psi wide confidence interval. This involves calculating the required Z-score that results in a confidence interval width of 30 psi.

  4. What sample size is needed to estimate μ with a precision of ±15 psi at a 95% confidence level? For a given margin of error, we can use the formula for sample size: n=(zα/2×σE)2n = \left( \frac{z_{\alpha/2} \times \sigma}{E} \right)^2 where EE is the margin of error (15 psi).


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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-scores
Confidence Intervals
Sample Mean
Statistical Inference

Formulas

Z = (X - μ) / σ
Confidence Interval = μ ± Z * (σ / √n)
Sample Size Calculation = (Z * σ / E)²

Theorems

Central Limit Theorem
Z-distribution for Normal Distribution
Confidence Interval Formula

Suitable Grade Level

University Level (Undergraduate)