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

By Selva Prabhakaran · Published April 20, 2026 · Last updated April 20, 2026

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.

RHeights with d p q r prefixes

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.

RExercise: density at seventy two inches

# 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

RDensity at seventy two solution

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.

RExercise: cumulative probability at seventy five

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

Click to reveal solution

RCumulative at seventy five solution

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.

RExercise: ninetieth percentile of heights

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

Click to reveal solution

RNinetieth percentile solution

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).

RSAT scores between six and seven hundred

# 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.

RExercise: IQ between ninety and one ten

# 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

RIQ interval solution

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.

RExercise: commute probability upper tail

# 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

RUpper tail commute solution

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.

RExercise: commute threshold at five percent

# 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

RCommute threshold solution

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.

RFree throws exact and at most

# 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.

RExercise: exactly three heads in ten tosses

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

Click to reveal solution

RExactly three heads solution

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.

RExercise: at most four defectives

# 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

RAt most four defectives solution

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.

RCall centre exact and upper tail

# 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.

RExercise: exactly four emails

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

Click to reveal solution

RExactly four emails solution

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.

RExercise: more than seven events

# 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

RMore than seven events solution

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.

Rt critical and chi squared critical

# 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.

RExercise: t critical at ninety nine percent

# 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

Rt critical solution

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.

RExercise: chi squared critical at one percent

# 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

RChi squared critical solution

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.

RExercise: confidence interval for sample mean

# 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

RConfidence interval solution

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).

RExercise: power of one sided z test

# 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

RPower of z test solution

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?

REnd-to-end bolt factory quality check

# 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`