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Statistics and Probability - Normal distribution

Grade 11IB_AA

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

🔑Concepts

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The Normal Distribution is a continuous probability distribution defined by two parameters: the mean μ\mu and the variance σ2\sigma^2, denoted as X∼N(μ,σ2)X \sim N(\mu, \sigma^2).

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The probability density function is bell-shaped and perfectly symmetric about the mean x=μx = \mu. At this point, the mean, median, and mode are identical.

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The total area under the normal curve is equal to 11. The curve is asymptotic to the horizontal axis (it never touches the xx-axis).

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The 'Empirical Rule' states that for a normal distribution: approximately 68%68\% of data falls within μ±σ\mu \pm \sigma, 95%95\% falls within μ±2σ\mu \pm 2\sigma, and 99.7%99.7\% falls within μ±3σ\mu \pm 3\sigma.

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The Standard Normal Distribution is a specific normal distribution where μ=0\mu = 0 and σ=1\sigma = 1, denoted as Z∼N(0,1)Z \sim N(0, 1).

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A zz-score represents the number of standard deviations a value xx is from the mean. It is used to compare different normal distributions by 'standardizing' them.

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Inverse Normal calculations are used to find a specific value kk when the probability (area) P(X<k)P(X < k) is already known.

📐Formulae

X∼N(μ,σ2)X \sim N(\mu, \sigma^2)

z=x−μσz = \frac{x - \mu}{\sigma}

P(a<X<b)=P(a−μσ<Z<b−μσ)P(a < X < b) = P\left(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}\right)

P(X>μ)=P(X<μ)=0.5P(X > \mu) = P(X < \mu) = 0.5

💡Examples

Problem 1:

The weights of bags of rice are normally distributed with a mean of 505505 g and a standard deviation of 1010 g. Find the probability that a randomly chosen bag weighs less than 490490 g.

Solution:

Let XX be the weight of a bag, so X∼N(505,102)X \sim N(505, 10^2). We need to find P(X<490)P(X < 490). Standardizing the value: z=490−50510=−1510=−1.5z = \frac{490 - 505}{10} = \frac{-15}{10} = -1.5 Using a GDC or ZZ-tables for P(Z<−1.5)P(Z < -1.5): P(X<490)≈0.0668P(X < 490) \approx 0.0668

Explanation:

First, identify the parameters μ=505\mu = 505 and σ=10\sigma = 10. Then, convert the raw score to a zz-score to determine how many standard deviations 490490 is from the mean. Finally, use the cumulative normal distribution function to find the area to the left of that zz-score.

Problem 2:

Given X∼N(μ,σ2)X \sim N(\mu, \sigma^2). If P(X<20)=0.0228P(X < 20) = 0.0228 and P(X>35)=0.1587P(X > 35) = 0.1587, find the values of μ\mu and σ\sigma.

Solution:

Step 1: Find zz-scores for both probabilities. For P(X<20)=0.0228P(X < 20) = 0.0228, the inverse normal ZZ-score is z1=−2.00z_1 = -2.00. For P(X>35)=0.1587P(X > 35) = 0.1587, then P(X<35)=1−0.1587=0.8413P(X < 35) = 1 - 0.1587 = 0.8413. The inverse normal ZZ-score is z2=1.00z_2 = 1.00. Step 2: Set up simultaneous equations using x=μ+zσx = \mu + z\sigma:

  1. 20=μ−2σ20 = \mu - 2\sigma
  2. 35=μ+1σ35 = \mu + 1\sigma Subtracting (1) from (2): 35−20=μ−μ+σ−(−2σ)35 - 20 = \mu - \mu + \sigma - (-2\sigma) 15=3σ  ⟹  σ=515 = 3\sigma \implies \sigma = 5 Substitute σ=5\sigma = 5 into (2): 35=μ+5  ⟹  μ=3035 = \mu + 5 \implies \mu = 30

Explanation:

When both μ\mu and σ\sigma are unknown, use the Inverse Normal function on a calculator to find the zz-scores corresponding to the given probabilities. Create a system of linear equations using the standardization formula and solve for the variables.

Problem 3:

The heights of students in a school are normally distributed with mean 170170 cm and standard deviation 88 cm. The tallest 10%10\% of students are invited to join the basketball team. What is the minimum height required to be invited?

Solution:

Let X∼N(170,82)X \sim N(170, 8^2). We want to find the value kk such that P(X>k)=0.10P(X > k) = 0.10. This is equivalent to P(X<k)=1−0.10=0.90P(X < k) = 1 - 0.10 = 0.90. Using the Inverse Normal function on a GDC: invNorm(area=0.90, mean=170, sd=8) k≈180.25k \approx 180.25 cm.

Explanation:

This is an Inverse Normal problem. Since the 'tallest 10%10\%' refers to the right tail of the distribution, we must find the value kk where the area to the left is 0.900.90 (or use the right-tail setting on a GDC).