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Statistics and Probability - Introduction to Probability Distributions (Binomial and Normal)

Grade 11IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Random Variables: A random variable is a numerical description of the outcome of a statistical experiment. Discrete random variables, like those in a Binomial distribution, take on countable values (e.g., 0,1,20, 1, 2). Continuous random variables, like those in a Normal distribution, can take any value within an interval. Visually, discrete variables are often represented by distinct points or vertical bars on a graph, while continuous variables are represented by a smooth, unbroken line or curve.

The Binomial Distribution XB(n,p)X \sim B(n, p): This distribution models the number of successes in a fixed number of independent trials nn, where each trial has only two possible outcomes (success or failure) and the probability of success pp remains constant. Visually, the distribution appears as a series of vertical bars (a histogram); if p=0.5p = 0.5, the bars are perfectly symmetrical around the center, but if p<0.5p < 0.5, the distribution is 'right-skewed' with a longer tail of bars extending to the right.

The Normal Distribution XN(μ,σ2)X \sim N(\mu, \sigma^2): A continuous probability distribution characterized by its 'bell-shaped' curve. The curve is perfectly symmetrical around the mean μ\mu, which is the highest point of the graph. The total area under the curve is exactly 11, representing the total probability. Visually, the curve approaches the horizontal axis at both ends but never touches it, showing that extreme values have very low but non-zero probabilities.

Standard Deviation and Curve Spread: In a Normal distribution, the standard deviation σ\sigma dictates the 'fatness' or 'skinniness' of the bell curve. A small σ\sigma results in a tall, narrow curve where data points are clustered closely around the mean μ\mu. A large σ\sigma creates a short, wide, and flatter curve, indicating that the data is more spread out from the center.

The Empirical Rule (68-95-99.7 Rule): For any Normal distribution, approximately 68%68\% of the data falls within 11 standard deviation of the mean (μ±σ\mu \pm \sigma), 95%95\% falls within 22 standard deviations (μ±2σ\mu \pm 2\sigma), and 99.7%99.7\% falls within 33 standard deviations (μ±3σ\mu \pm 3\sigma). Visually, this is shown by shading sections of the bell curve: the middle section containing 68%68\% of the area is the most dense, while the 'tails' beyond 3σ3\sigma contain very little area (0.3%0.3\% combined).

Standard Normal Distribution and Z-scores: The Standard Normal Distribution is a special case where the mean μ=0\mu = 0 and standard deviation σ=1\sigma = 1, denoted as ZN(0,1)Z \sim N(0, 1). A Z-score represents how many standard deviations a value xx is from the mean. On a graph, a positive Z-score is located to the right of the center, and a negative Z-score is to the left. This allows us to compare values from different normal distributions on a single standardized scale.

Cumulative Probability: For both distributions, the cumulative probability P(Xk)P(X \le k) represents the probability that the variable takes a value less than or equal to kk. In a Normal distribution, this is represented visually as the total shaded area under the curve to the left of the vertical line x=kx = k. For a Binomial distribution, it is the sum of the heights of all bars from x=0x = 0 up to x=kx = k.

📐Formulae

Binomial Probability: P(X=r)=(nr)pr(1p)nrP(X = r) = \binom{n}{r} p^r (1-p)^{n-r}

Binomial Coefficient: (nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!(n-r)!}

Expected Value (Mean) of Binomial: E(X)=npE(X) = np

Variance of Binomial: Var(X)=np(1p)Var(X) = np(1-p)

Z-score Formula: z=xμσz = \frac{x - \mu}{\sigma}

Normal Distribution Notation: XN(μ,σ2)X \sim N(\mu, \sigma^2)

Standardization: P(X<x)=P(Z<xμσ)P(X < x) = P(Z < \frac{x - \mu}{\sigma})

💡Examples

Problem 1:

A biased coin has a probability of 0.60.6 of landing on heads. If the coin is flipped 88 times, find the probability of getting exactly 55 heads.

Solution:

Step 1: Identify the parameters. This is a Binomial distribution XB(n,p)X \sim B(n, p) where n=8n = 8 and p=0.6p = 0.6. We want to find P(X=5)P(X = 5). Step 2: Use the Binomial formula: P(X=5)=(85)(0.6)5(10.6)85P(X = 5) = \binom{8}{5} (0.6)^5 (1 - 0.6)^{8-5}. Step 3: Calculate the coefficient: (85)=8!5!3!=8×7×63×2×1=56\binom{8}{5} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. Step 4: Calculate the powers: (0.6)5=0.07776(0.6)^5 = 0.07776 and (0.4)3=0.064(0.4)^3 = 0.064. Step 5: Multiply the values: P(X=5)=56×0.07776×0.0640.2787P(X = 5) = 56 \times 0.07776 \times 0.064 \approx 0.2787.

Explanation:

We identify the problem as Binomial because there are a fixed number of trials, constant probability, and independent flips. We plug the values into the P(X=r)P(X=r) formula to find the specific probability for 55 successes.

Problem 2:

The masses of apples in a harvest are normally distributed with a mean of 150150 g and a standard deviation of 1212 g. Find the probability that a randomly selected apple has a mass less than 135135 g.

Solution:

Step 1: Define the distribution: XN(150,122)X \sim N(150, 12^2). We want to find P(X<135)P(X < 135). Step 2: Calculate the Z-score for x=135x = 135: z=13515012=1512=1.25z = \frac{135 - 150}{12} = \frac{-15}{12} = -1.25. Step 3: Use the standard normal table or a calculator to find P(Z<1.25)P(Z < -1.25). Step 4: P(Z<1.25)0.1056P(Z < -1.25) \approx 0.1056.

Explanation:

To solve problems involving the Normal distribution, we first standardize the value using the Z-score formula. This tells us that 135135 g is 1.251.25 standard deviations below the mean. The area to the left of this Z-score on the standard normal curve gives us the required probability.