R Probability Distributions Exercises: 12 dnorm/pnorm/qnorm Problems — Solved Step-by-Step)

R's d/p/q/r convention — dnorm, pnorm, qnorm, rnorm — lets you pull a density, a cumulative probability, a quantile, or a random sample from any distribution using one naming pattern. This 12-problem set drills that convention across the normal, binomial, Poisson, t, and chi-squared distributions, with a starter, click-to-reveal solution, and plain-language explanation for every problem.

How do `d`, `p`, `q`, and `r` prefixes work in R distributions?

Every distribution in R ships with four functions that share a prefix-plus-name pattern: d for density, p for the cumulative probability, q for the quantile (inverse of p), and r for random samples. Swap the name — norm, binom, pois, t, chisq — and the same four prefixes work. That single idea powers every problem below. Watch all four act on one example: adult male heights, modelled as normal with mean 69 inches and standard deviation 3 inches.

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mu <- 69 sigma <- 3 # Density at height = 70 inches dnorm(70, mu, sigma) #> [1] 0.1258 # Cumulative: P(height <= 75) pnorm(75, mu, sigma) #> [1] 0.9772 # Quantile: 95th percentile of heights qnorm(0.95, mu, sigma) #> [1] 73.9346 # Random: 5 simulated heights set.seed(42) rnorm(5, mu, sigma) #> [1] 73.11 67.31 70.09 70.90 70.21

The density 0.1258 is the height of the bell curve at 70 inches — not a probability (it's a rate per inch). The cumulative 0.9772 says 97.72% of men are 75 inches or shorter. The 95th percentile 73.93 is the height below which 95% of men fall. The five random draws simulate men pulled at random from the same bell curve. Same four prefixes — they work on every distribution R knows.

Key Insight

Memorise the prefixes once and you've memorised every distribution. d = density (or mass, for discrete), p = cumulative probability up to a value, q = the value at a given cumulative probability, r = random sample. The distribution name follows: norm, binom, pois, t, chisq, and dozens more.

Problem 1: Density value from `dnorm`

Try it: For the same height distribution N(69, 3), compute the density at 72 inches using dnorm(). Store the result in p1_density.

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# Problem 1: dnorm at x = 72, mu = 69, sigma = 3 p1_density <- # your code here p1_density #> Expected: roughly 0.0807

Click to reveal solution

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p1_density <- dnorm(72, mean = 69, sd = 3) p1_density #> [1] 0.0807

Explanation: dnorm(x, mean, sd) returns the value of the probability density function at x. 0.0807 is the curve's height at 72 inches — a density, not a probability. To turn a density into a probability you integrate it over an interval, which is exactly what pnorm() does in the next problem.

Problem 2: Cumulative probability from `pnorm`

Try it: What fraction of men are 75 inches or shorter on the same N(69, 3) distribution? Store the answer in p2_prob.

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# Problem 2: P(height <= 75) p2_prob <- # your code here p2_prob #> Expected: roughly 0.9772

Click to reveal solution

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p2_prob <- pnorm(75, mean = 69, sd = 3) p2_prob #> [1] 0.9772

Explanation: pnorm(q, mean, sd) is the cumulative distribution function — the area under the bell curve to the left of q. 0.9772 says 97.72% of men are 75 inches or shorter, which means 75 sits at about 2 standard deviations above the mean (the classic "68-95-99.7 rule" again).

Problem 3: Quantile from `qnorm`

Try it: Find the 90th percentile of heights on N(69, 3) — the height below which 90% of men fall. Store it in p3_quantile.

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# Problem 3: 90th percentile of N(69, 3) p3_quantile <- # your code here p3_quantile #> Expected: roughly 72.84

Click to reveal solution

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p3_quantile <- qnorm(0.90, mean = 69, sd = 3) p3_quantile #> [1] 72.8446

Explanation: qnorm(p, mean, sd) is the inverse of pnorm() — give it a cumulative probability and it returns the x-value. pnorm(72.8446, 69, 3) would give you back 0.90. The q-functions are how you translate percentile language ("top 10%", "bottom quartile") into actual numbers.

How do I compute probabilities for non-standard normal distributions?

Most real distributions aren't standard normal — they have their own mean and spread. The same pnorm() handles them with no extra work; you just pass mean and sd. For a range like "between 600 and 700", subtract two pnorm() calls. For upper tails like "at least 45 minutes", use lower.tail = FALSE — it's more numerically stable than 1 - pnorm(...) when you're far out in the tail. Here's the between-pattern on SAT scores modelled as N(500, 100).

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# SAT scores: N(500, 100). What fraction scored between 600 and 700? sat_between <- pnorm(700, 500, 100) - pnorm(600, 500, 100) sat_between #> [1] 0.1359

pnorm(700, 500, 100) gives the area to the left of 700 (≈ 0.9772), and pnorm(600, 500, 100) gives the area to the left of 600 (≈ 0.8413). Subtract and you get the strip between them: 13.59% of test-takers scored in that range. The same subtract-two-pnorms pattern works for any "between a and b" question on any distribution that has a cumulative function.

Tip

Use lower.tail = FALSE for upper-tail probabilities, not 1 - pnorm(...). In the far right tail 1 - pnorm(x) can round to zero because of floating-point subtraction, while pnorm(x, lower.tail = FALSE) computes the tail directly and stays accurate. Habit counts — always reach for lower.tail = FALSE.

Problem 4: Between two values on a normal distribution

Try it: IQ scores are modelled as N(100, 15). What fraction of people have IQs between 90 and 110? Store the answer in p4_prob.

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# Problem 4: P(90 <= IQ <= 110) on N(100, 15) p4_prob <- # your code here p4_prob #> Expected: roughly 0.4950

Click to reveal solution

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p4_prob <- pnorm(110, 100, 15) - pnorm(90, 100, 15) p4_prob #> [1] 0.4950

Explanation: 90 and 110 sit symmetrically around the mean of 100, each about two-thirds of a standard deviation away. Just under half the population (≈ 49.5%) falls in that band. The "68-95-99.7 rule" applies at exact integer multiples of σ; for fractional multiples you let R do the integration.

Problem 5: Upper tail with `lower.tail = FALSE`

Try it: Commute times are modelled as N(35, 8) minutes. What fraction of commutes last longer than 45 minutes? Use lower.tail = FALSE. Store the answer in p5_prob.

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# Problem 5: P(commute > 45) on N(35, 8) using lower.tail = FALSE p5_prob <- # your code here p5_prob #> Expected: roughly 0.1056

Click to reveal solution

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p5_prob <- pnorm(45, mean = 35, sd = 8, lower.tail = FALSE) p5_prob #> [1] 0.1056

Explanation: 45 minutes sits 1.25 standard deviations above the mean of 35. lower.tail = FALSE flips pnorm from "area to the left" to "area to the right", giving the upper-tail probability directly. Roughly 10.56% of commutes run longer than 45 minutes.

Problem 6: Finding the top-5% threshold with `qnorm`

Try it: For the same commute distribution N(35, 8), find the commute duration above which only the slowest 5% of commutes fall. Store the answer in p6_threshold.

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# Problem 6: threshold T such that P(commute > T) = 0.05 p6_threshold <- # your code here p6_threshold #> Expected: roughly 48.16

Click to reveal solution

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p6_threshold <- qnorm(0.95, mean = 35, sd = 8) p6_threshold #> [1] 48.1588

Explanation: "Top 5%" means "above the 95th percentile". qnorm(0.95, ...) returns the value below which 95% of commutes fall — and therefore above which the remaining 5% fall. You could equivalently write qnorm(0.05, 35, 8, lower.tail = FALSE); both return the same 48.16 minutes.

How do I compute binomial probabilities with `dbinom` and `pbinom`?

The binomial distribution models the count of successes in a fixed number of independent trials, each with the same success probability. The R parameters are size (number of trials) and prob (success probability). Because the counts are discrete, dbinom returns an actual probability — P(X = k) — not a density. pbinom(k, ...) returns P(X ≤ k), inclusive of k. Here's a basketball free-throw example with ten attempts at 70% accuracy.

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# 10 free throws, p = 0.7 # P(exactly 8 made) and P(at most 5 made) ft_exact8 <- dbinom(8, size = 10, prob = 0.7) ft_atmost5 <- pbinom(5, size = 10, prob = 0.7) c(ft_exact8, ft_atmost5) #> [1] 0.2335 0.1503

dbinom(8, 10, 0.7) gives the exact probability of making exactly 8 of 10 shots: 23.35%. pbinom(5, 10, 0.7) gives the probability of making 5 or fewer: 15.03%. Notice how the "at most 5" is much less likely than "exactly 8" — because 7 is the expected number of makes, counts below the mean quickly become rare. The same dbinom/pbinom pair works for A/B test conversions, defective-rate sampling, and any other success-count setting.

Note

pbinom(k, ...) is P(X ≤ k), not P(X < k). It includes k itself. For strict inequality you need pbinom(k - 1, ...). This trips people up because the continuous pnorm() doesn't have this ambiguity — with continuous distributions P(X = k) = 0, so ≤ and < agree. With discrete distributions they don't.

Problem 7: Exact count with `dbinom`

Try it: A fair coin is tossed 10 times. What is the probability of exactly 3 heads? Store the answer in p7_prob.

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# Problem 7: P(exactly 3 heads in 10 tosses) p7_prob <- # your code here p7_prob #> Expected: roughly 0.1172

Click to reveal solution

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p7_prob <- dbinom(3, size = 10, prob = 0.5) p7_prob #> [1] 0.1172

Explanation: There are $\binom{10}{3} = 120$ ways to get exactly 3 heads out of 10 tosses, and each specific sequence has probability $0.5^{10} = 1/1024$. Multiply: $120/1024 = 0.1172$. dbinom() does that arithmetic for you and generalises to any prob, not just 0.5.

Problem 8: Cumulative count with `pbinom`

Try it: A factory produces widgets with a 10% defective rate. In a batch of 20, what is the probability of finding 4 or fewer defectives? Store the answer in p8_prob.

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# Problem 8: P(at most 4 defectives in batch of 20, p = 0.10) p8_prob <- # your code here p8_prob #> Expected: roughly 0.9568

Click to reveal solution

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p8_prob <- pbinom(4, size = 20, prob = 0.10) p8_prob #> [1] 0.9568

Explanation: With a 10% defective rate, the expected number of defectives in 20 widgets is just 2, so seeing 4 or fewer is very likely — about 95.68%. The remaining 4.32% is the probability of seeing 5 or more defectives. This is the building block of acceptance-sampling plans in quality control.

How do I model rare events with the Poisson distribution?

The Poisson distribution models counts of events that occur independently at a known average rate per unit time (or area, or volume). Its single parameter lambda is both the mean and the variance. R uses dpois for exact counts and ppois for cumulative. Here's a call-centre example at 12 calls per hour.

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# Call centre: lambda = 12 calls/hour # P(exactly 15 calls) and P(more than 20 calls) cc_exact15 <- dpois(15, lambda = 12) cc_morethan20 <- ppois(20, lambda = 12, lower.tail = FALSE) c(cc_exact15, cc_morethan20) #> [1] 0.0724 0.0116

dpois(15, 12) tells us that in any given hour there's a 7.24% chance of exactly 15 calls — plausible but above the mean of 12. ppois(20, 12, lower.tail = FALSE) gives the probability of a burst: more than 20 calls in an hour, about 1.16%. The same pattern handles defect counts per production shift, goals per football match, or cosmic rays per square centimetre per second — anywhere events arrive at a steady average rate.

Problem 9: Exact count with `dpois`

Try it: You receive on average 3 work emails per hour. Assuming a Poisson arrival process, what is the probability of receiving exactly 4 emails in the next hour? Store the answer in p9_prob.

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# Problem 9: P(exactly 4 emails) when lambda = 3 p9_prob <- # your code here p9_prob #> Expected: roughly 0.1680

Click to reveal solution

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p9_prob <- dpois(4, lambda = 3) p9_prob #> [1] 0.1680

Explanation: Plugging into the Poisson pmf $P(X = k) = e^{-\lambda} \lambda^k / k!$ gives $e^{-3} \cdot 3^4 / 24 \approx 0.168$. R spares you the arithmetic. Note that 4 is just above the mean of 3, so this probability is near the peak of the distribution.

Problem 10: Upper tail with `ppois` and `lower.tail = FALSE`

Try it: A town records on average 5 traffic accidents per week. What is the probability of more than 7 accidents in any given week? Store the answer in p10_prob.

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# Problem 10: P(X > 7) when lambda = 5, using lower.tail = FALSE p10_prob <- # your code here p10_prob #> Expected: roughly 0.1334

Click to reveal solution

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p10_prob <- ppois(7, lambda = 5, lower.tail = FALSE) p10_prob #> [1] 0.1334

Explanation: ppois(7, 5) would give P(X ≤ 7) ≈ 0.8666. Flipping to lower.tail = FALSE gives the complement directly: P(X > 7) ≈ 0.1334 — roughly a 1-in-7.5 chance of a bad week. That's the number an insurer or a public-health analyst plans against.

How do I find critical values for t and chi-squared distributions?

Hypothesis tests and confidence intervals need critical values — cutoffs that separate the typical from the extreme under a null distribution. The q-functions return them directly. qt(p, df) gives t critical values for any degrees of freedom, and qchisq(p, df) does the same for chi-squared. For a two-sided CI at confidence level level, pass (1 + level) / 2 as the probability — you want the tail (1 - level)/2 on each side.

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# t critical for two-sided 95% CI, df = 24 t_crit_95 <- qt(0.975, df = 24) # chi-squared critical for alpha = 0.05 (upper tail), df = 10 chisq_crit_05 <- qchisq(0.95, df = 10) c(t_crit_95, chisq_crit_05) #> [1] 2.0639 18.3070

The t critical 2.0639 says: a 95% CI on a mean with 24 degrees of freedom extends 2.0639 standard errors either side of the sample mean. The chi-squared critical 18.307 is the right-tail cutoff — any chi-squared statistic above it rejects at α = 0.05 in a one-sided test. These two q-calls replace a lookup table and generalise to any df.

Warning

q-functions return lower-tail quantiles by default. For a two-sided CI at 95% confidence, use qt(0.975, df) — not qt(0.95, df). The probability you pass is the cumulative area to the left, so you want (1 + 0.95)/2 = 0.975 to leave 2.5% in each tail. Getting this off-by-one wrong silently builds too-narrow intervals.

Problem 11: t critical value for a 99% CI

Try it: Find the t critical value for a two-sided 99% confidence interval with 15 degrees of freedom. Store it in p11_crit.

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# Problem 11: t critical for two-sided 99% CI, df = 15 p11_crit <- # your code here p11_crit #> Expected: roughly 2.9467

Click to reveal solution

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p11_crit <- qt(0.995, df = 15) p11_crit #> [1] 2.9467

Explanation: For 99% confidence, leave 0.5% in each tail — so pass (1 + 0.99)/2 = 0.995 to qt(). A 99% CI is wider than the 95% CI at the same df (2.95 vs 2.13), which is the cost of buying more confidence. The t critical also grows as df shrinks, reflecting the extra uncertainty of small samples.

Problem 12: Chi-squared critical value for α = 0.01

Try it: Find the chi-squared critical value for an upper-tail test at α = 0.01 with 8 degrees of freedom. Store it in p12_crit.

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# Problem 12: chi-squared critical for alpha = 0.01, df = 8 p12_crit <- # your code here p12_crit #> Expected: roughly 20.0902

Click to reveal solution

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p12_crit <- qchisq(0.99, df = 8) p12_crit #> [1] 20.0902

Explanation: Chi-squared tests (goodness-of-fit, independence, variance) are one-sided — large values reject. For α = 0.01 you want the 99th percentile of the chi-squared distribution with 8 df, which qchisq(0.99, 8) returns as 20.09. Any observed statistic above 20.09 rejects the null at the 1% level.

Practice Exercises

These two capstone problems combine multiple concepts from the tutorial above. Solve them end-to-end before revealing the answer.

Exercise 1: Build a 95% confidence interval for a sample mean

Simulate 30 observations from N(50, 10) with set.seed(2026). Compute a 95% confidence interval for the mean using the sample mean, sample standard deviation, and a t critical value. Store the lower and upper bounds in cap1_ci as a numeric vector of length 2.

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# Capstone 1: 95% CI for a sample mean set.seed(2026) cap1_sample <- rnorm(30, mean = 50, sd = 10) # Hint: use mean(), sd(), qt(), and length() cap1_ci <- # your code here cap1_ci #> Expected: roughly c(46.4, 53.3)

Click to reveal solution

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set.seed(2026) cap1_sample <- rnorm(30, mean = 50, sd = 10) cap1_mean <- mean(cap1_sample) cap1_se <- sd(cap1_sample) / sqrt(length(cap1_sample)) cap1_crit <- qt(0.975, df = length(cap1_sample) - 1) cap1_ci <- c(cap1_mean - cap1_crit * cap1_se, cap1_mean + cap1_crit * cap1_se) cap1_ci #> [1] 46.3839 53.2661

Explanation: The formula is $\bar{x} \pm t_{0.975, n-1} \cdot s/\sqrt{n}$. Three R ingredients: mean() for the centre, sd()/sqrt(n) for the standard error, and qt(0.975, n-1) for the t critical value. The simulated mean drifts slightly from 50 (the true mean) and the CI covers it — which it should, 95% of the time.

Exercise 2: Power of a one-sided z-test

Consider a one-sided z-test of $H_0: \mu = 100$ vs $H_1: \mu > 100$ with $\sigma = 15$ (known), $n = 25$, and $\alpha = 0.05$. Compute the power of the test when the true mean is $\mu = 104$. Store the answer in cap2_power.

The recipe:

Find the rejection threshold $\bar{x}^{*}$ on the sample-mean scale under $H_0$ using qnorm().
Compute the probability that the sample mean exceeds $\bar{x}^{*}$ under $H_1$ using pnorm(..., lower.tail = FALSE).

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# Capstone 2: power of a one-sided z-test sigma <- 15 n <- 25 mu0 <- 100 mu1 <- 104 alpha <- 0.05 cap2_power <- # your code here cap2_power #> Expected: roughly 0.3777

Click to reveal solution

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sigma <- 15 n <- 25 mu0 <- 100 mu1 <- 104 alpha <- 0.05 se <- sigma / sqrt(n) x_crit <- qnorm(1 - alpha, mean = mu0, sd = se) cap2_power <- pnorm(x_crit, mean = mu1, sd = se, lower.tail = FALSE) cap2_power #> [1] 0.3777

Explanation: The standard error of the sample mean is $\sigma/\sqrt{n} = 3$. Under $H_0$, we reject when the sample mean exceeds qnorm(0.95, 100, 3) ≈ 104.93. Under $H_1$ ($\mu = 104$), the probability of exceeding 104.93 is pnorm(104.93, 104, 3, lower.tail = FALSE) ≈ 0.3777. So the test has only a 38% chance of detecting the real 4-unit effect with n = 25 — an argument for a larger sample size.

Complete Example

End-to-end scenario combining three of the skills from this tutorial. A factory produces bolts whose diameters are normally distributed with mean 10 mm and standard deviation 0.1 mm. The spec window is 9.8–10.2 mm. Management asks three questions: (a) what fraction of bolts are in-spec, (b) if we sample 50 bolts, what's the probability that none are out of spec, and (c) at what diameter does the upper 1% of bolts start?

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# Factory diameters: N(10, 0.1). Spec window: 9.8 to 10.2. mu_d <- 10 sigma_d <- 0.1 # (a) Fraction in-spec qc_in_spec <- pnorm(10.2, mu_d, sigma_d) - pnorm(9.8, mu_d, sigma_d) # (b) P(zero defectives in a sample of 50) p_defective <- 1 - qc_in_spec qc_zero_def <- dbinom(0, size = 50, prob = p_defective) # (c) Diameter above which only 1% of bolts fall qc_cutoff <- qnorm(0.99, mu_d, sigma_d) c(in_spec = qc_in_spec, zero_def_50 = qc_zero_def, top1_cutoff = qc_cutoff) #> in_spec zero_def_50 top1_cutoff #> 0.9545 0.0975 10.2326

Step (a) computes the strip between 9.8 and 10.2 by subtracting two pnorm() calls — 95.45% of bolts land inside spec. Step (b) reuses that result: the defective rate is 1 − 0.9545 = 4.55%, and dbinom(0, 50, 0.0455) gives the chance of a perfect batch of 50 as about 9.75% — surprisingly low, because with any non-trivial defect rate a batch of 50 almost always has at least one bad unit. Step (c) uses qnorm(0.99, ...) to find the upper-1% cutoff at 10.23 mm — a useful threshold for screening outliers. Three distributions, three prefixes, one coherent pipeline.

Summary

The d/p/q/r prefix convention is the single most portable pattern in R statistical computing. Memorise it on the normal distribution and it transfers to every other distribution with no new rules.

Prefix	Returns	Given	Typical question
`d`	Density (continuous) or probability mass (discrete)	`x`, parameters	"Height of the curve at x"
`p`	Cumulative probability	`q`, parameters	"What fraction of X ≤ q?"
`q`	Quantile (inverse CDF)	`p`, parameters	"What x has p probability below it?"
`r`	Random sample	`n`, parameters	"Give me n draws from this distribution"

The five distribution families covered here — norm, binom, pois, t, chisq — take different parameter names (mean/sd, size/prob, lambda, df) but share the same four prefixes. Reach for lower.tail = FALSE for upper tails, remember pbinom/ppois are inclusive (P(X ≤ k)), and convert CI confidence levels to quantile probabilities with (1 + level) / 2.

For the full mechanics of the t, F, normal, and chi-squared distributions — which situations generate each one and how to read their shapes — see the parent tutorial Normal, t, F, and Chi-Squared Distributions in R.

References

R Core Team — Distributions manual page (?Distributions). Link
R Core Team — The Normal Distribution reference (?Normal). Link
R Core Team — The Binomial Distribution reference (?Binomial). Link
R Core Team — The Poisson Distribution reference (?Poisson). Link
R Core Team — The Student t Distribution reference (?TDist). Link
Verzani, J. — Using R for Introductory Statistics (2nd ed.), Chapter 5: Probability and Distributions. CRC Press (2014).
Kross, S. — Introduction to dnorm, pnorm, qnorm, and rnorm for new biostatisticians. Link

Continue Learning

Normal, t, F, and Chi-Squared Distributions in R — the parent core post covering each distribution's shape, parameters, and when it arises.
Probability in R Exercises — a broader sibling exercise set covering simulation, Bayes, and classical probability.
Binomial vs Poisson in R: Understand When Each Distribution Applies — choosing between the two most common discrete distributions.

R Probability Distributions Exercises: 12 dnorm/pnorm/qnorm Problems — Solved Step-by-Step)

How do d, p, q, and r prefixes work in R distributions?

Problem 1: Density value from dnorm

Problem 2: Cumulative probability from pnorm

Problem 3: Quantile from qnorm

How do I compute probabilities for non-standard normal distributions?

Problem 4: Between two values on a normal distribution

Problem 5: Upper tail with lower.tail = FALSE

Problem 6: Finding the top-5% threshold with qnorm

How do I compute binomial probabilities with dbinom and pbinom?

Problem 7: Exact count with dbinom

Problem 8: Cumulative count with pbinom

How do I model rare events with the Poisson distribution?

Problem 9: Exact count with dpois

Problem 10: Upper tail with ppois and lower.tail = FALSE

How do I find critical values for t and chi-squared distributions?

Problem 11: t critical value for a 99% CI

Problem 12: Chi-squared critical value for α = 0.01

Practice Exercises

Exercise 1: Build a 95% confidence interval for a sample mean

Exercise 2: Power of a one-sided z-test

Complete Example

Summary

References

Continue Learning

Related Tutorials

How do `d`, `p`, `q`, and `r` prefixes work in R distributions?

Problem 1: Density value from `dnorm`

Problem 2: Cumulative probability from `pnorm`

Problem 3: Quantile from `qnorm`

Problem 5: Upper tail with `lower.tail = FALSE`

Problem 6: Finding the top-5% threshold with `qnorm`

How do I compute binomial probabilities with `dbinom` and `pbinom`?

Problem 7: Exact count with `dbinom`

Problem 8: Cumulative count with `pbinom`

Problem 9: Exact count with `dpois`

Problem 10: Upper tail with `ppois` and `lower.tail = FALSE`