Normal Distribution - Statistics

• This chapter is about the normal distribution
• It is for 2nd Semester, 2024–2025

Learning Outcomes

• Discuss the importance of the normal distribution in Statistics
• Define the characteristics of the normal curve
• Compute and interpret z scores
• Use a table of areas under the normal curve to solve problems involving the normal distribution
• Use the central limit theorem to solve problems involving sample means for large samples

Normal and Standard Normal Distribution

• Probability distributions can be discrete or continuous, leading to different types of distributions
• Discrete Probability Distribution includes:
  • Uniform distribution
  • Binomial distribution
  • Poisson distribution
• Continuous Probability Distribution includes the Normal distribution

Normal Distribution

• The normal probability distribution is statistically useful and important
• It's an assumption of almost all statistical tests used for hypothesis testing
• The graph of a random variable X that follows a normal distribution is called the normal curve
• Normal Curve Properties
  • The mode, point on the horizontal axis where the maximum curve occurs, occurs at x = μ
  • The curve is symmetric about a vertical axis through the mean μ
  • Curve approaches the horizontal axis asymptotically in either direction away from the mean
  • The total area under the curve and above the horizontal axis equals 1
• The equation for the probability distribution of the normal random variable depends on μ and σ²
  • The density function calculation for a normal random variable is:
  • f(x) = 1/(√(2πσ²)) * exp[-(1/2) * ((x-μ)/σ)²]
  • X can take up values from negative to positive infinity
  • Values of the normal density function can be written as X~N (μ, σ²)
• For the equationX~N(μ, σ²), a normal random variable has mean μ, and variance σ²

Standard Normal Distribution

• Calculating the probability of a normal random variable X directly can be complex
• Standardizing the scores of normal random variable X simplifies probability calculations
• Standard normal distribution refers to a normal probability distribution with a mean of 0, a standard deviation of 1, with a total area under its density curve of 1
• It's denoted by random variable Z and is written as Z~N(0, 1)
• Determining the probability of a normal random variable standardizes X-scale scores to Z-scores
• x to z transformation uses the equation Z = (χ - μ) / σ

Finding Area Under the Standard Normal Curve

• Calculate the area under the curve (or probability) of standard normal random variables
  • Draw a normal curve; shade the area
  • Search for a Z value in the standard normal table
  • Perform the math to compute the probability of the shaded area
• The values in the table represent the area or probability corresponding to the Z-score
• The table shows the probability from Z-score = 0 to any point
• Z-scores range from -3 to 3
• The probability is the same when Z-scores are negative
• Example of how to find probabilties :
  • To find the probability greater Z=0.42 then the calculation is
• P(Z > 0.42) = 0.5 - 0.1628 = 0.3372
  • To find the probability greater Z=-0.42 then the calculation is:
• P(Z > -0.42) = 0.5 + 0.1628 = 0.6628
  • To find the probability between Z=-0.42 and Z=1.11 then the calculation is:
• P(-0.42 < Z < 1.11) = P(Z = -0.42) + P(Z = 1.11) = 0.1628 + 0.3665 = 0.5293
  • To find the probability between Z=-0.42 and Z=-1.11 then the calculation is:
• P(−1.11 < Z < -0.42) = P(Z = -1.11) - P(Z = -0.42) = 0.3665 - 0.1628 = 0.2037

Continuous Distribution Summary

• Normal Distribution
  • Random Variable X: continuous
  • Possible Values of X: {-∞, +∞}
  • Parameter: μ, σ²
  • Probability Function f(x): f(x) = (1/√(2πσ²)) * exp[-1/2 * ((x - μ) / σ)²]
  • Mean: μ
  • Variance: σ²
• Standard Normal
  • Random Variable X: continuous
  • Possible Values of X: {-∞, +∞}
  • Parameter: μ, σ²
  • Probability Function f(x): Z = (X - μ) / σ
  • Mean: 0
  • Variance: 1

Application of Normal Distribution

• If IQ scores are normally distributed with mean 100 and standard deviation 20, calculating the probability of a person having an IQ score of at least 130 will be calculated as:
  • The calculation begins with the transformation P(X ≥ 130) = P(Z ≥ (130-100)/20) = P(Z ≥ 1.5) = 0.5-0.4332 = 0.0668
  • The probability of having an IQ of at least 130 is 6.68%
• If the average number of calories in a 1.5-ounce chocolate bar is 225, and the distribution of calories is approximately normal with σ2 = 100, find the probability that a randomly selected chocolate bar will have between 200 and 235 calories:
  • The calculation begins with the transformation P(200 ≤ X ≤ 235) = P((200-225)/10 ≤ Z ≤ (235-225)/10) = P(−2.5 ≤ Z ≤ 1) = P(Z = -2.5) + P(Z = 1) = 0.4938 + 0.3413 = 0.8351
  • The probability that a randomly selected chocolate bar will have between 200 and 235 calories is 83.51%

The Central Limit Theorem

• As the sample size n increases without limit, the shape of the sample means taken with replacement from a population with mean μ and standard deviation σ will approach a normal distribution
• The distribution will have a mean μ and a standard deviation º/√n"
  • Sample Means Approximation: X ~N(μ, σ²/n)
  • Z Score Calculation: Z = (X - μ) / (σ/√n)

Remarks

• The distribution of the sample means will be normally distributed for any sample size n for any normally distributed original variable
• When the variable distribution is not normal and the sample size n is > 30, the distribution of the sample means can be approximated relatively well by a normal distribution
• The approximation improves as the sample size n grows
• When the average time spent by students studying on weekdays is 7.93 hours, the distribution is normal, and the standard deviation is 0.8 hours, calculating the probability that the mean of the sample will be greater than 8 hours for a sample of 40 students is as follows:
  • Apply the central limit theorem: P(X > 8) = P(Z > (8-7.93) / (0.8/√40)) = P(Z > 0.55) = 0.5 - 0.2088 = 0.2912
  • The probability that the mean of the samples will be greater than 8 hours is calculated to be 29.12%
• The average person uses 123 gallons of water daily and the standard deviation is 21 gallons, find the probability that the mean of a randomly selected sample of 15 people will be between 115 and 135 gallons, assuming the variable is normally distributed:
  • Formula Application: P(115 ≤ X ≤ 135) = P((115-123) / (21/√15) ≤ Z ≤ (135-123) / (21/√15)) = P(-1.48 ≤ Z ≤ 2.21) = P(Z = -1.48) + P(Z = 2.21) = 0.4306 + 0.4864 = 0.917
  • The probability of the mean for a sample of 15 being selected between 115-135 gallons of water is 91.7%