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Math III · Mathematics
Probability · The normal or Gaussian distribution

The Normal Distribution and the Normal Approximation to the Binomial

Before you start, make sure you know: Intervals, half-lines and neighborhoods (Integrated Math II)
Before you start, make sure you know: Linear equations (Math 8 (Pre-Algebra))

A machine makes 600 parts, and 5% of them come out defective. What is the probability that a box has between 25 and 35 defective parts? This is a binomial situation, but adding up eleven terms like $\binom{600}{k}(0.05)^k(0.95)^{600-k}$ by hand would take forever. When $n$ is large, the binomial histogram looks like a bell curve, so we can use the normal distribution instead. First we need to master the normal curve, then we'll see how to use it for binomials.

The Normal Curve

Heights of 11th graders, the battery life of a phone, the weight of apples from one orchard: when you measure many individuals, the data often pile up in the middle and thin out symmetrically on both sides. That shape is the normal distribution.

A bigger $\sigma$ gives a wider, flatter bell. Because $X$ is continuous, $P(X = a) = 0$ for any single value $a$. Only intervals have probability.

Normal curve with the central bands within 1, 2 and 3 standard deviations of the mean, containing about 68%, 95% and 99.7% of the area. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% 95% 99.7%

Play with the sliders: the first one moves the center $\mu$, the second one changes the spread $\sigma$. Notice that the area under the curve stays equal to $1$.

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The weight of apples from a harvest follows a normal distribution with a mean of $150$ ounces and a standard deviation of $15$ ounces. If apples weighing between $135$…

The weight of apples from a harvest follows a normal distribution with a mean of $150$ ounces and a standard deviation of $15$ ounces. If apples weighing between $135$ and $165$ ounces are considered "type A apples", calculate the probability that a randomly chosen apple is type A.

σ=15 σ=15 Type A 135 150 165 μ-σ μ μ+σ
Solve

z-Scores: Standardizing a Value

Suppose Mia scores 27 on the ACT and Jayden scores 1250 on the SAT. Who did better? You can't compare 27 and 1250 directly, because the tests use different scales. The fix is to ask: how many standard deviations is each score from its own mean?

Standardizing: any normal variable X with mean mu and standard deviation sigma is converted to the standard normal variable Z. X ~ N(μ, σ) Z ~ N(0, 1) Z = (X − μ)/σ A z-value tells how many standard deviations X is above (Z > 0) or below (Z < 0) the mean.

Standardizing is the key trick of this whole page. Every normal distribution, whatever its units, becomes the same $N(0,1)$, so one table is enough for all of them.

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Two friends, Marcus and Sarah, study at different schools. At Marcus's school, the mean score on the Economics exam was $6,2$ with a standard deviation of $1,1$. At…

Two friends, Marcus and Sarah, study at different schools. At Marcus's school, the mean score on the Economics exam was $6,2$ with a standard deviation of $1,1$. At Sarah's, the mean was $6,8$ with a standard deviation of $1,5$. If Marcus obtained a $7,5$ and Sarah a $8,2$, and assuming that both populations follow a normal distribution, determine which of the two obtained a better relative score compared to their group.

3 4 5 6 7 8 9 10 Score μ₁=6,2 μ₂=6,8 Marcus (7.5) Sarah (8.2) +1,18σ₁ +0,93σ₂ Marcus: σ₁ = 1.1 Sarah: σ₂ = 1.5
Solve

The Standard Normal Table

The table lists the cumulative area to the left of $z$, that is, $P(Z \le z)$. Find the row for the ones digit and tenths digit of $z$, then the column for the hundredths digit. For example, $z = 1.32$ is row $1.3$, column $0.02$, which gives $0.9066$.

$z$ 0.00 0.01 0.02 0.03 0.04 0.05
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115
Standard normal curve with the area to the left of a value z shaded; this is the area the table gives. P(Z ≤ z) 0 z
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Given the standard normal distribution $Z \sim N(0, 1)$, use the probability table to calculate the cumulative area to the left of the following values. Graphically…

Given the standard normal distribution $Z \sim N(0, 1)$, use the probability table to calculate the cumulative area to the left of the following values. Graphically represent the situation in part a.

  1. a) $P(Z \le 1,32)$
  2. b) $P(Z < 2,05)$
  3. c) $P(Z \le 0,48)$
Z −3 −2 −1 1 2 3 0 1,32 Area = P(Z ≤ 1.32)
Solve
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In a $Z \sim N(0, 1)$ distribution, we know that the total area under the curve is equal to $1$. Use the properties of symmetry and the complementary event to find the…

In a $Z \sim N(0, 1)$ distribution, we know that the total area under the curve is equal to $1$. Use the properties of symmetry and the complementary event to find the following probabilities:

  1. a) $P(Z > 1,15)$
  2. b) $P(Z \ge 0,64)$
  3. c) $P(Z > 2,80)$
Solve

Probabilities in Real Situations

When a problem gives $\mu$ and $\sigma$, the recipe is always the same.

Normal curve for daily minutes of video games with the area to the right of 150 minutes shaded, corresponding to z = 1.6. Video games: X ~ N(110, 25) minutes. Find P(X > 150). 110 150 X = 150 → Z = (150 − 110)/25 = 1.6
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A study on the time 16-year-old teenagers spend using a well-known social network per day indicates that it follows a normal distribution with a mean of $\mu = 120$…

A study on the time 16-year-old teenagers spend using a well-known social network per day indicates that it follows a normal distribution with a mean of $\mu = 120$ minutes and a standard deviation of $\sigma = 20$ minutes.

If we choose a young person at random, calculate the probability that they use the social network for less than $150$ minutes. To do this, standardize the variable $X$ and use the $N(0, 1)$ table.

X Z μ=120 150 σ=20 P(X < 150) 0 z = ? Standardize
Solve
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The scores of an Applied Mathematics for Social Sciences exam follow a $N(5,2; 1,5)$ distribution. If a score higher than $8$ is required to obtain a special mention,…

The scores of an Applied Mathematics for Social Sciences exam follow a $N(5,2; 1,5)$ distribution. If a score higher than $8$ is required to obtain a special mention, what is the probability of a student achieving it?

Hint: Remember that $P(X > k) = 1 - P(X \le k)$ after standardizing.

Solve
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The time that high school juniors dedicate daily to doing homework follows a normal distribution with a mean of $\mu = 115$ minutes and a standard deviation of…

The time that high school juniors dedicate daily to doing homework follows a normal distribution with a mean of $\mu = 115$ minutes and a standard deviation of $\sigma = 25$ minutes. Calculate the probability that a randomly chosen student:

  1. a) Dedicates between 90 and 140 minutes a day.
  2. b) Dedicates more than 160 minutes.
Time (min) 115 μ 90 140 160 σ=25 a) b) N(115, 25)
Solve

Working Backward: Finding a Cutoff Value

Sometimes you know the area and need the value: "What score puts you in the top 25%?" Now you read the table backward.

Normal curve with the top 25 percent tail shaded to find the cutoff value z0. area 0.75 0.25 μ x top 25%: z₀ ≈ 0.67, x = μ + 0.67σ

Here are some $z$-values you'll use often. Each one is the $z$ whose left area is the number shown.

Area to the left 0.80 0.85 0.90 0.95 0.975 0.99
$z$ 0.84 1.04 1.28 1.645 1.96 2.33
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Find the value of the abscissa $k$ (the value of the variable on the horizontal axis) in a standard normal distribution $Z \sim N(0, 1)$ knowing that: a)…

Find the value of the abscissa $k$ (the value of the variable on the horizontal axis) in a standard normal distribution $Z \sim N(0, 1)$ knowing that:

  1. a) $P(Z \le k) = 0,8413$
  2. b) $P(Z \le k) = 0,9750$
  3. c) $P(Z \le k) = 0,5000$
Solve
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A gaming peripherals brand has determined that the response time of its flagship mouse follows a $N(18, 4)$ milliseconds distribution. The marketing department wants to…

A gaming peripherals brand has determined that the response time of its flagship mouse follows a $N(18, 4)$ milliseconds distribution. The marketing department wants to promote that their product is so fast that only $15\%$ of competitors' mice (which follow the same distribution) exceed a critical response time $k$. Find the value of this time $k$.

Solve
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The battery life of a smartphone model follows a $N(42, 6)$ hours distribution. The company wants to offer a replacement warranty if the battery lasts less than a…

The battery life of a smartphone model follows a $N(42, 6)$ hours distribution. The company wants to offer a replacement warranty if the battery lasts less than a certain time. What maximum duration should the company guarantee if it is only willing to replace $2,5\%$ of the batteries sold?

μ = 42 k σ = 6 2,5%
Solve

Middle Intervals

A "central" interval leaves equal tails on both sides. If the middle holds $c$ of the area, each tail holds $\dfrac{1 - c}{2}$, so the area to the left of the upper endpoint is $\dfrac{1 + c}{2}$.

Normal curve with the central 95 percent shaded between mu minus 1.96 sigma and mu plus 1.96 sigma, and 2.5 percent in each tail. 95% 2.5% 2.5% μ − 1.96σ μ + 1.96σ
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In a prestigious esports tournament, the score obtained by the participants follows a normal distribution with a mean of 2500 points and a standard deviation of 400…

In a prestigious esports tournament, the score obtained by the participants follows a normal distribution with a mean of 2500 points and a standard deviation of 400 points. The organization decides to award the "Elite Mention" only to the 2.5% of players with the highest scores.

  1. a) From what score is this mention received?
  2. b) If a group of friends has scores that lie in the central interval comprising 95% of the players in the tournament, between which two values do their scores range?
Solve

Finding $\mu$ and $\sigma$ from Probabilities

Sometimes the unknown is the distribution itself. Each probability you're given turns into a $z$-value, and each $z$-value gives an equation $z_0 = \dfrac{x - \mu}{\sigma}$. With one unknown parameter you need one equation. With both unknown you get a system of two linear equations, which you can solve by elimination.

Normal curve with a left tail of 2.28 percent below 40 and a right tail of 15.87 percent above 70, used to find the mean and standard deviation. 2.28% 15.87% 40 μ = ? 70 z = −2 at 40 and z = 1 at 70
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It is known that the weight of coffee packets from a certain factory follows a normal distribution. After a quality control check, it is observed that $5\%$ of the…

It is known that the weight of coffee packets from a certain factory follows a normal distribution. After a quality control check, it is observed that $5\%$ of the packets weigh less than $242$ ounces and $10\%$ weigh more than $255$ ounces. Find the mean $\mu$ and the standard deviation $\sigma$ of this population.

Solve
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A courier company uses drones for its deliveries in urban centers. It has been proven that the flight time of these drones until the battery runs out follows a normal…

A courier company uses drones for its deliveries in urban centers. It has been proven that the flight time of these drones until the battery runs out follows a normal distribution with a mean of 42 minutes. If 97.72% of the drones fly for more than 30 minutes, what is the probability that a randomly chosen drone runs out of battery between 35 and 45 minutes of flight?

Solve
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It is known that the charging time of the batteries of a new model of electric car follows a normal distribution $N(\mu, \sigma)$. After carrying out a market study, it…

It is known that the charging time of the batteries of a new model of electric car follows a normal distribution $N(\mu, \sigma)$. After carrying out a market study, it is observed that 15% of the vehicles take less than 45 minutes to charge completely, while only 20% of the vehicles require more than 75 minutes.

Calculate the value of the mean $\mu$ and the standard deviation $\sigma$ of this distribution. Then, determine the probability that a randomly chosen vehicle takes between 50 and 60 minutes to charge.

15% 20% ? 45 75 μ 50 60 X ~ N(μ, σ)
Solve

When Can a Binomial Be Approximated by a Normal?

Now back to binomials. If $X \sim B(n, p)$ counts successes in $n$ independent trials, then $\mu = np$ and $\sigma = \sqrt{npq}$, where $q = 1 - p$. As $n$ grows, the histogram of $X$ looks more and more like a bell.

Histogram of a binomial distribution with a smooth bell curve of the normal distribution drawn over it. B(n, p) with large n ≈ bell curve of N(np, √(npq))

The approximation fails when $p$ is close to $0$ or $1$ and $n$ isn't big enough, because the histogram is lopsided.

Once you know $\mu$ and $\sigma$, finding a probability is exactly the same $z$-score work as before.

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A basketball free-throw shooter has a probability of success of $p = 0,8$. If they take $n = 100$ shots, we want to know if we can approximate the binomial variable $X$…

A basketball free-throw shooter has a probability of success of $p = 0,8$. If they take $n = 100$ shots, we want to know if we can approximate the binomial variable $X$ (number of successful shots) by a normal distribution.

  1. a) Check if the conditions $n \cdot p \ge 5$ and $n \cdot q \ge 5$ are met.
  2. b) If so, state what the parameters $\mu$ and $\sigma$ of the new normal distribution $X'$ would be.
Solve
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A fair coin ($p = 0,5$) is flipped $400$ times. Calculate the probability of getting fewer than $190$ heads using the normal approximation to the binomial (without…

A fair coin ($p = 0,5$) is flipped $400$ times. Calculate the probability of getting fewer than $190$ heads using the normal approximation to the binomial (without applying the continuity correction yet).

Step 1: Calculate the mean $\mu$ and the standard deviation $\sigma$.

Step 2: Standardize the value $X = 190$ and look it up in the table.

Solve

The Continuity Correction

A binomial takes only whole-number values ($0, 1, 2, \dots$), but a normal variable is continuous. For a normal, $P(X' = 65) = 0$. To fix this, we treat each whole number $k$ as a bar of width $1$, spanning from $k - 0.5$ to $k + 0.5$. Some problems call this Yates's continuity correction.

Continuity correction: the integer bar for k becomes the continuous interval from k minus 0.5 to k plus 0.5. the integer bar of the binomial becomes this continuous stretch [k − 0.5, k + 0.5] k k − 0.5 k + 0.5
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On a streaming platform, $15\%$ of users cancel their subscription before the end of the trial month. If $400$ new users have signed up this month, calculate, using the…

On a streaming platform, $15\%$ of users cancel their subscription before the end of the trial month. If $400$ new users have signed up this month, calculate, using the normal approximation with continuity correction, the probability that exactly $65$ users unsubscribe.

Solve
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A multiple-choice exam consists of $80$ questions, each with $4$ answer options of which only one is correct. If a student answers all the questions completely at…

A multiple-choice exam consists of $80$ questions, each with $4$ answer options of which only one is correct. If a student answers all the questions completely at random:

  1. a) Determine the parameters of the normal distribution that approximates this variable.
  2. b) Calculate the probability that they get at least $25$ correct answers.
Solve

Solving Complete Problems

You now have all the pieces.

Normal curve with mean 60 and standard deviation 6 and the area between 54.5 and 70.5 shaded. 54.5 60 70.5 P(54.5 ≤ X′ ≤ 70.5), with μ = 60 and σ = 6
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A study on recycling habits states that $35\%$ of households in a city separate plastic correctly. If a random sample of $120$ households is selected: a) Justify why…

A study on recycling habits states that $35\%$ of households in a city separate plastic correctly. If a random sample of $120$ households is selected:

  1. a) Justify why this experiment can be approximated by a normal distribution.
  2. b) Calculate the probability that, in this sample, between $40$ and $50$ households (inclusive) separate plastic.
Solve
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A study on reading habits in teenagers reveals that 70% of young people aged between 15 and 16 prefer digital format over paper. If we select a random sample of 400…

A study on reading habits in teenagers reveals that 70% of young people aged between 15 and 16 prefer digital format over paper. If we select a random sample of 400 teenagers:

  1. a) Justify whether it is possible to approximate the distribution of the variable "number of teenagers who prefer digital format" using a normal distribution.
  2. b) Calculate, using the previous approximation and applying the continuity correction, the probability that between 270 and 290 teenagers (inclusive) prefer the digital format.
READING HABITS IN TEENAGERS p = 0.70 (Digital) q = ? (Paper) Random sample: n = 400 a) Is it possible to approximate to the Normal? n · p = ? (≥ 5) n · q = ? (≥ 5) X ~ B(400, 0.7) ➔ N(μ = ?, σ = ?) b) Probability with continuity correction Standardization: Z = (X - μ) / σ μ = ? σ = ? 270 290 269.5 290.5 P ( 269.5 ≤ X' ≤ 290.5 ) = ?
Solve
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A biased coin is flipped 500 times, where the probability of getting heads is 0.4. Calculate the probability of getting: a) Exactly 200 heads. b) More than 215 heads. c)…

A biased coin is flipped 500 times, where the probability of getting heads is 0.4. Calculate the probability of getting:

  1. a) Exactly 200 heads.
  2. b) More than 215 heads.
  3. c) Fewer than 185 heads.

Note: Use the normal approximation with Yates's continuity correction.

Solve

Harder Problems: Typical Ranges, Cutoffs, and the Correction

The interval $np \pm 2\sigma$

Because $X'$ is normal, about 95% of the time $X$ lands within two standard deviations of $np$. This tells you which outcomes are "typical" and which would be surprising.

Normal curve with the central region from np minus 2 sigma to np plus 2 sigma shaded, covering about 95 percent of expected results. np − 2σ np np + 2σ about 95% of typical results

Finding a cutoff for a binomial

To find a count that a given probability cuts off, solve for $x$ with $x = \mu + z_0\sigma$ exactly as before. Then undo the continuity correction. We want the smallest whole number $k$ with $P(X \le k) \ge$ the given probability. Since $P(X \le k) \to P(X' \le k + 0.5)$, we need $k + 0.5 \ge x$.

Does the correction really matter?

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5% of the parts manufactured by a high-precision machine are defective. If they are packed in boxes of 200 units, calculate the probability that in a box: a) There are…

5% of the parts manufactured by a high-precision machine are defective. If they are packed in boxes of 200 units, calculate the probability that in a box:

  1. a) There are exactly 10 defective parts.
  2. b) There are more than 15 defective parts.
  3. c) The number of defective parts is less than or equal to 5.

Argue whether the results would be significantly different if the continuity correction were not applied.

Solve
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In a multiple-choice exam with 100 questions, each with 4 answer options (only one correct), a student answers all the questions at random. a) What is the probability…

In a multiple-choice exam with 100 questions, each with 4 answer options (only one correct), a student answers all the questions at random.

  1. a) What is the probability that they get more than 30 questions right?
  2. b) If at least 40 correct answers are needed to pass the exam, what is the probability that the student passes purely by chance?
  3. c) Determine the interval $(np - 2\sigma, np + 2\sigma)$ and interpret its meaning in this context.
10 15 20 25 30 35 40 X ~ B(100, 0.25) ≈ N(25, 4.33) μ = 25 a) P(X > 30) b) P(X ≥ 40) 16.34 33.66 c) μ ± 2σ Contains ~95.4% of the probability
Solve
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A gym chain knows that 20% of its registered members never visit the facilities. If there are 600 registered members in a specific gym: a) Calculate the probability that…

A gym chain knows that 20% of its registered members never visit the facilities. If there are 600 registered members in a specific gym:

  1. a) Calculate the probability that the number of members who never visit is between 110 and 135 inclusive.
  2. b) If the gym decides to launch a motivation campaign for these members if the number of "inactive" members exceeds 140, what is the probability that they will have to launch this campaign?
  3. c) What is the maximum number of inactive members expected to be found with a probability of 90%?
Solve

Back to the start: For the machine with $X \sim B(600, 0.05)$, $np = 30 \ge 5$ and $nq = 570 \ge 5$, so $\mu = 30$ and $\sigma = \sqrt{28.5} \approx 5.34$. With the correction, $P(25 \le X \le 35) \to P(24.5 \le X' \le 35.5) = P(-1.03 \le Z \le 1.03) = 2(0.8485) - 1 = 0.6970$. There's about a 69.7% chance that a box has between 25 and 35 defective parts.

Summary

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