Stats class notes 10/29

Introduction to Probability Distributions

This section covers various topics related to probability distributions, specifically focusing on normal distributions and the calculations involved in determining probabilities and percentiles for specified random variables based on provided parameters. It includes examples related to various datasets.

Part A: Chocolate Chips in Chips Ahoy! Cookies

Context: The number of chocolate chips in an 18-ounce bag of Chips Ahoy! cookies follows an approximately normal distribution with:

  • Mean ($BC$): 1262 chips

  • Standard Deviation ($C3$): 118 chips

  • Source: Brad Warner and Jim Rutledge, Chance 12(1): 10-14, 1999

Problem A1: Probability Calculation

(a) Calculate the probability that a randomly selected bag contains between 1000 and 1400 chocolate chips, inclusive.

Method: Use the normal distribution formula to find the probabilities for $X = 1000$ and $X = 1400$.
Apply the Z-score formula:
Z = rac{X - BC}{C3}

Probabilities:

  • For $X = 1000$:
    Z1000=rac1000−1262118<br>em−2.2186Z_{1000} = rac{1000 - 1262}{118} <br>em -2.2186
    Using a standard normal distribution table, find the area under the curve for this Z-value.

  • For $X = 1400$:
    Z1400=rac1400−1262118<br>em1.166Z_{1400} = rac{1400 - 1262}{118} <br>em 1.166

The probability of being between these two Z-scores results in the calculation of the area between them in the standard normal distribution, denoted as $P(1000 ext{ ≤ } X ext{ ≤ } 1400)$.

Problem A2: Probability of Fewer Chips

(b) Calculate the probability that a randomly selected bag contains fewer than 1000 chocolate chips.

This requires the Z-score calculation for 1000 as shown before and finding the area to the left of this Z-score in the standard normal distribution.

Problem A3: Proportion of Bags

(c) Determine the proportion of bags containing more than 1200 chocolate chips.

First, compute the Z-score:

  • For $X = 1200$:
    Z1200=rac1200−1262118<br>em−0.5254Z_{1200} = rac{1200 - 1262}{118} <br>em -0.5254
    And find the corresponding area in the standard normal distribution to the right of this Z-score.

Problem A4: Probability of Fewer Chips than 1125

(d) Find the proportion of bags containing fewer than 1125 chocolate chips.

Calculate the Z-score for 1125:

  • Z1125=rac1125−1262118<br>em−1.161Z_{1125} = rac{1125 - 1262}{118} <br>em -1.161
    Then calculate the corresponding area.

Problem A5: Percentile Rank for 1475 Chips

(e) Calculate the percentile rank of a bag that contains 1475 chocolate chips.

For 1475, calculate:

  • Z1475=rac1475−1262118<br>em1.798Z_{1475} = rac{1475 - 1262}{118} <br>em 1.798
    Using standard normal tables or calculators determine $P(X ext{ ≤ } 1475)$, which helps find the percentile rank.

Result: 96.45 percentile.

Problem A6: Percentile Rank for 1050 Chips

(f) Calculate the percentile rank of a bag containing 1050 chocolate chips.

Calculate:

  • Z1050=rac1050−1262118<br>em−1.794Z_{1050} = rac{1050 - 1262}{118} <br>em -1.794
    Determine the associated area to find the percentile rank.

Part B: House Prices and Normal Distributions

Problem B1: Price Range for Middle 80%

A contractor aims to build houses that appeal to the middle 80% of buyers. Given:

  • Mean house price ($BC$): $325,000

  • Standard Deviation ($C3$): $12,000

Goal: Find the range of prices such that:
P(Lowerext≤Xext≤Upper)=0.80P(Lower ext{ ≤ } X ext{ ≤ } Upper) = 0.80

To find the limits, determine the Z-scores that correspond to the percentiles at 10% and 90% for the normal distribution. The result will give the price range for the proposed houses to meet the needs of the targeted buyers.

Results for Range:

  • Lower Limit: $309,621.38

  • Upper Limit: $340,378.62

Part C: Sampling Distribution

Problem C1: IQ Score Distribution

An Intelligence Quotient (IQ) measurement is derived from tests that are approximately normally distributed with:

  • Mean ($BC$): 100

  • Standard Deviation ($C3$): 15

To find the sampling distribution of the sample mean for a sample size of n = 9, the following must be considered:

  • The sampling distribution will have the same mean as the population:
    BC_{ar{X}} = BC

  • The standard deviation of the sample means will be:
    C3_{ar{X}} = rac{C3}{
    ho{n}} = rac{15}{
    ho{9}} = 5

Central Limit Theorem

  • It states that the distribution of the sample means approaches a normal distribution as the sample size increases, regardless of the population distribution.

Sampling Distribution for the Sample Proportion

  • The conditions of a binomial distribution require:

    1. np ≥ 10

    2. n(1-p) ≥ 10

    3. Sample size n must be less than 5% of the population N.

This ensures the validity of using normal approximation for proportion calculations.