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Statistics and Probability - Normal distribution

Grade 11IB_AI

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

🔑Concepts

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A continuous random variable XX follows a Normal Distribution if its probability density function is bell-shaped and symmetrical about the mean μ\mu. It is denoted as X∼N(μ,σ2)X \sim N(\mu, \sigma^2), where μ\mu is the mean and σ2\sigma^2 is the variance.

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Properties of the Normal Curve: The mean, median, and mode are all equal. The total area under the curve is 11. The curve is asymptotic to the horizontal axis.

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The Empirical Rule (68-95-99.7 Rule): Approximately 68%68\% of the data lies within one standard deviation of the mean (μ±σ)(\mu \pm \sigma), 95%95\% within two standard deviations (μ±2σ)(\mu \pm 2\sigma), and 99.7%99.7\% within three standard deviations (μ±3σ)(\mu \pm 3\sigma).

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Standard Normal Distribution: This is a specific normal distribution where the mean is 00 and the standard deviation is 11. It is denoted as Z∼N(0,1)Z \sim N(0, 1).

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The zz-score represents the number of standard deviations a value xx is from the mean μ\mu.

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Inverse Normal Distribution: Used to find a value xx (or kk) when a probability (area) is given, such that P(X<k)=pP(X < k) = p. On a GDC, this is the 'invNorm' function.

📐Formulae

z=x−μσz = \frac{x - \mu}{\sigma}

X∼N(μ,σ2)X \sim N(\mu, \sigma^2) where μ\mu is the mean and σ\sigma is the standard deviation.

P(x1<X<x2)=P(x1−μσ<Z<x2−μσ)P(x_1 < X < x_2) = P\left(\frac{x_1 - \mu}{\sigma} < Z < \frac{x_2 - \mu}{\sigma}\right)

💡Examples

Problem 1:

The weights of bags of rice are normally distributed with a mean of 5.025.02 kg and a standard deviation of 0.050.05 kg. Find the probability that a randomly selected bag weighs less than 5.005.00 kg.

Solution:

X∼N(5.02,0.052)X \sim N(5.02, 0.05^2) We need to find P(X<5.00)P(X < 5.00). Using the GDC (normCDF with lower bound −∞-\infty or −1099-10^{99}, upper bound 5.005.00, μ=5.02\mu = 5.02, σ=0.05\sigma = 0.05): P(X<5.00)≈0.344578...P(X < 5.00) \approx 0.344578... P(X<5.00)≈0.345 (3 s.f.)P(X < 5.00) \approx 0.345 \text{ (3 s.f.)}

Explanation:

Identify the distribution parameters μ=5.02\mu = 5.02 and σ=0.05\sigma = 0.05. Use the cumulative normal distribution function on the calculator to find the area to the left of 5.005.00.

Problem 2:

The heights of students in a school are normally distributed with μ=170\mu = 170 cm and σ=8\sigma = 8 cm. If the tallest 15%15\% of students are allowed to join the basketball team, what is the minimum height required to join?

Solution:

X∼N(170,82)X \sim N(170, 8^2) We want to find hh such that P(X>h)=0.15P(X > h) = 0.15. This is equivalent to P(X<h)=1−0.15=0.85P(X < h) = 1 - 0.15 = 0.85. Using the inverse normal function (invNorm with area 0.850.85, μ=170\mu = 170, σ=8\sigma = 8): h=invNorm(0.85,170,8)h = \text{invNorm}(0.85, 170, 8) h≈178.291...h \approx 178.291... Minimum height ≈178\approx 178 cm (to 3 s.f.)

Explanation:

Since the calculator 'invNorm' function typically requires the area to the left, we subtract 15%15\% from 100%100\% to get 0.850.85. We then find the value of hh corresponding to this cumulative probability.

Problem 3:

Given X∼N(20,σ2)X \sim N(20, \sigma^2) and P(X<25)=0.8P(X < 25) = 0.8, find the standard deviation σ\sigma.

Solution:

First, find the zz-score for which P(Z<z)=0.8P(Z < z) = 0.8 using Z∼N(0,1)Z \sim N(0, 1). Using GDC: z=invNorm(0.8,0,1)≈0.8416z = \text{invNorm}(0.8, 0, 1) \approx 0.8416 Using the zz-score formula: z=x−μσz = \frac{x - \mu}{\sigma} 0.8416=25−20σ0.8416 = \frac{25 - 20}{\sigma} 0.8416=5σ0.8416 = \frac{5}{\sigma} σ=50.8416≈5.941...\sigma = \frac{5}{0.8416} \approx 5.941... σ≈5.94 (3 s.f.)\sigma \approx 5.94 \text{ (3 s.f.)}

Explanation:

Standardize the variable using the zz-score formula. Since we don't know σ\sigma, we use the Standard Normal Distribution to find the zz-value that corresponds to the given probability, then solve for σ\sigma.