Skip to content
NumeriveMATH
Make learning comfortable
Changing style, motion, or reading speed keeps your current question, answers, homework, and saved progress. The school band still chooses your task layout. Saved to your learner profile across devices. All lessons and help stay available.
← University Probability & Statistics levels

University Probability & Statistics / LEVEL 1 · DIFFICULTY 4/5

University Probability & Statistics

Read the prerequisite, study a worked example and justify each step in these selected university topics.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE USTAT 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Random Variables, PMFs, PDFs & CDFs

Useful preparation: Conditional Sample Spaces

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A random variable turns an uncertain outcome into a number. Its distribution assigns probabilities to sets of possible numbers. For isolated outcomes we add masses; for a variable with a density we integrate area. The cumulative distribution function unifies both descriptions by asking how much probability has accumulated at or below a chosen threshold.

Probability mass function
For a discrete random variable X, p (x)=P (X=x)\displaystyle p\, \left(x\right) = P\, \left(X = x\right). Its values are nonnegative and sum to one over the possible values.
Probability density function
A density f is nonnegative and integrates to one. Integrating f over an interval gives its probability; f(x) itself is not the probability of one point.
Cumulative distribution function
F (x)=P\displaystyle F\, \left(x\right) = P(X at most x) is nondecreasing and right-continuous, approaching zero and one at the two infinite ends. It exists for every real-valued random variable.

Conditions before calculation

  • A proposed mass or density must be nonnegative everywhere on its support and have total probability one. State the support and give zero mass or density outside it.
  • A binomial count requires a fixed number n of independent trials, each with the same success probability p and two possible classifications. Sampling without replacement generally violates independence.
  • For a variable with a density, each single point has probability zero. For any distribution, P(a<X\displaystyle a < X at most b)=F (b)−F (a)\displaystyle F\, \left(b\right) - F\, \left(a\right); changing endpoint inclusion can matter for discrete distributions.

Why the method works

1. Choose the correct model

Countable outcome values suggest a mass function. Measurements modeled continuously may use a density. Mixed distributions can have both jumps and continuous portions, so do not force every distribution into one pure type.

2. Normalize before calculating

The total area or sum must equal one. A density may exceed one if its support is narrow enough; only integrated probabilities must remain between zero and one.

3. Use the CDF to subtract sets

Removing outcomes at or below a from outcomes at or below b leaves precisely the half-open interval.

P(a<X≤b)=F(b)−F(a)P(a<X\le b)=F(b)-F(a)

Worked examples

EXAMPLE 1

Three independent sensors each signal with probability 0.2. Find the probability that exactly one signals.

  1. Let X count signals. The fixed trial count, independence, and common probability give a binomial model.

  2. There are three choices for the successful sensor. Each specific pattern has the same probability.

    P(X=1)=(31)(0.2)(0.8)2P(X=1)=\binom31(0.2)(0.8)^2

  3. Multiply to get the mass; the four masses for counts zero through three also sum to one.

    P(X=1)=0.384,0.512+0.384+0.096+0.008=1P(X=1)=0.384,\quad 0.512+0.384+0.096+0.008=1

Conclusion: The probability is 0.384, provided the independence assumption is appropriate.

EXAMPLE 2

Let f (x)=c x\displaystyle f\, \left(x\right) = c\, x on [0,2] and zero elsewhere. Find c and P(1<X\displaystyle 1 < X at most 2).

  1. Normalize the density by integrating over its entire support.

    1=∫02cx dx=2c,c=121=\int_0^2 cx\,dx=2c,\qquad c=\frac12

  2. Integrate from zero to a threshold inside the support to get the CDF there.

    F(x)=x24(0≤x≤2)F(x)=\frac{x^2}{4}\quad(0\le x\le2)

  3. Subtract the CDF values; the endpoints carry no mass.

    P(1<X≤2)=F(2)−F(1)=1−14=34P(1<X\le2)=F(2)-F(1)=1-\frac14=\frac34

Conclusion: The probability is 34\displaystyle \frac{3}{4}. Complete the CDF with zero for x<0\displaystyle x < 0 and one for x>2\displaystyle x > 2.

Pitfalls & counterexamples

Tempting claim: A probability density cannot exceed one.

Correction: Density is probability per unit length, so only its area is bounded by one.

Test the claim: The uniform density on [0,0.2] equals 5 but has total area 5 times 0.2=1\displaystyle 0.2 = 1.

Check your understanding

If X has a density, what is P (X=1.5)\displaystyle P\, \left(X = 1.5\right)?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

Zero.

A single point has zero interval length; a positive density value does not create a point mass.

Connect this idea

Distributions supply the weights for expectation and variance. Their CDFs support quantiles, simulation, and probability calculations used in statistical inference.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    A discrete variable has outcomes 0,1,2 with probabilities 14\displaystyle \frac{1}{4}, 24\displaystyle \frac{2}{4}, and p. Find p.

    Practise this exact problem →
  2. Problem 2

    X is uniform on [0,6]. Find P (X≤3)\displaystyle P\, \left(X \le 3\right).

    Practise this exact problem →
  3. Problem 3

    Two independent trials each succeed with probability 12\displaystyle \frac{1}{2}. Find the probability both succeed.

    Practise this exact problem →
  4. Problem 4

    A continuous variable has CDF values F (2)=14\displaystyle F\, \left(2\right) = \frac{1}{4} and F (5)=34\displaystyle F\, \left(5\right) = \frac{3}{4}. Find P (2<X≤5)\displaystyle P\, \left(2 < X \le 5\right).

    Practise this exact problem →
  5. Problem 5New context

    A reward variable has outcomes 0,1,2 with probabilities 15\displaystyle \frac{1}{5}, 25\displaystyle \frac{2}{5}, and p. Find p.

    Practise this exact problem →
  6. Problem 6New context

    A waiting time X is uniform on [0,8]. Find P (X≤4)\displaystyle P\, \left(X \le 4\right).

    Practise this exact problem →
  7. Problem 7New context

    Two independent quality checks each succeed with probability 13\displaystyle \frac{1}{3}. Find the probability both succeed.

    Practise this exact problem →
  8. Problem 8New context

    A continuous sensor value has CDF values F (2)=15\displaystyle F\, \left(2\right) = \frac{1}{5} and F (5)=35\displaystyle F\, \left(5\right) = \frac{3}{5}. Find P (2<X≤5)\displaystyle P\, \left(2 < X \le 5\right).

    Practise this exact problem →
Open stage USTAT 1.1 in the student workspace →

STAGE USTAT 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Expectation, Variance & the CLT

Useful preparation: Random Variables, PMFs, PDFs & CDFs

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Expectation is a probability-weighted balance point, not a promise about any single observation. Variance measures squared distance from that balance point. Averaging independent observations preserves their common expectation but reduces their variability. The central limit theorem adds a statement about the shape of the distribution of the standardized average, under specific assumptions.

Expectation
The mean mu=E[X] is the sum of x times its probability mass, or the integral of x times its density, when the expectation exists. E[g(X)] weights g(x) using the same distribution.
Variance
Variance is E[(X-mu) squared], equivalently E[X squared]-mu squared when the second moment is finite. Standard deviation is its square root and has the original measurement units.
Standard error
The standard error of an estimator is its sampling standard deviation. For the mean of n independent identically distributed observations with variance sigma squared, it is sigma divided by the square root of n.

Conditions before calculation

  • Linearity of expectation does not require independence, but the relevant expectations must exist. Nonlinear functions generally cannot be moved outside expectation.
  • Variances of independent variables add. For dependent variables, covariance terms must be included; multiplying a random variable by a constant multiplies its variance by that constant squared.
  • The classical central limit theorem assumes independent identically distributed observations with finite positive variance. No fixed sample-size threshold guarantees a good normal approximation for every distribution.

Why the method works

1. Calculate moments with weights

First compute E[X], then E[X squared]. Squaring the mean after averaging is a different operation from averaging squared observations.

Var⁡(X)=E[X2]−(E[X])2\operatorname{Var}(X)=E[X^2]-(E[X])^2

2. Track scaling

Adding a constant shifts every outcome equally, leaving spread unchanged. Multiplying distances by a multiplies squared distances by a squared.

E[aX+b]=aE[X]+b,Var⁡(aX+b)=a2Var⁡(X)E[aX+b]=aE[X]+b,\qquad \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)

3. Distinguish exact facts from approximation

The mean and variance of the sample mean are exact under the stated assumptions. Its approximate normal shape is the additional asymptotic conclusion.

E[Xˉ]=μ,Var⁡(Xˉ)=σ2n,n(Xˉ−μ)σ→dN(0,1)E[\bar X]=\mu,\quad \operatorname{Var}(\bar X)=\frac{\sigma^2}{n},\quad \frac{\sqrt n(\bar X-\mu)}{\sigma}\xrightarrow{d}N(0,1)

Worked examples

EXAMPLE 1

X takes values 0, 2, and 5 with probabilities 0.2, 0.5, and 0.3. Find its mean and variance.

  1. The probabilities sum to one. Compute the weighted mean.

    E[X]=0(0.2)+2(0.5)+5(0.3)=2.5E[X]=0(0.2)+2(0.5)+5(0.3)=2.5

  2. Square outcomes before weighting to find the second moment.

    E[X2]=0+4(0.5)+25(0.3)=9.5E[X^2]=0+4(0.5)+25(0.3)=9.5

  3. Subtract the square of the mean.

    Var⁡(X)=9.5−2.52=3.25\operatorname{Var}(X)=9.5-2.5^2=3.25

Conclusion: The mean 2.5 need not be a possible observed value.

EXAMPLE 2

Independent repeated measurements have mean 40 and standard deviation 6. Find the spread of an average of 36 measurements, and describe a normal approximation.

  1. The average retains expectation 40 by linearity.

  2. Independence adds the 36 variances before division by 36 squared.

    Var⁡(Xˉ)=36⋅62362=1,SE⁡(Xˉ)=1\operatorname{Var}(\bar X)=\frac{36\cdot6^2}{36^2}=1,\qquad \operatorname{SE}(\bar X)=1

  3. If the normal approximation is adequate, standardizing 38 and 42 gives -2 and 2.

    P(38≤Xˉ≤42)≈P(−2≤Z≤2)≈0.9545P(38\le\bar X\le42)\approx P(-2\le Z\le2)\approx0.9545

Conclusion: The standard error is exactly 1; the probability approximation requires information about the underlying distribution and is exact in form if the measurements are normal.

Pitfalls & counterexamples

Tempting claim: The variance of 3X is three times the variance of X.

Correction: Variance scales by the square of the multiplier.

Test the claim: If X is equally likely to be -1 or 1, its variance is 1, whereas 3X has variance 9.

Check your understanding

Does increasing sample size make the original observations normally distributed?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

The central limit theorem concerns the standardized mean, not a change in the population distribution.

Connect this idea

Standard errors quantify sampling uncertainty in confidence intervals and tests. Dependence, heavy tails, and biased sampling can prevent familiar averaging formulas from answering the intended question.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    X equals 1 or 5, each with probability 12\displaystyle \frac{1}{2}. Find E[X].

    Practise this exact problem →
  2. Problem 2

    A random variable X has variance 2. Find Var(3 X+7\displaystyle 3\, X + 7).

    Practise this exact problem →
  3. Problem 3

    A population has standard deviation 6. For an independent random sample of size 4, what is the standard error of the sample mean?

    Practise this exact problem →
  4. Problem 4

    X is binomial n=8\displaystyle n = 8 and p=12\displaystyle p = \frac{1}{2}. Find Var(X).

    Practise this exact problem →
  5. Problem 5New context

    A fair reward X equals 2 or 6, each with probability 12\displaystyle \frac{1}{2}. Find E[X].

    Practise this exact problem →
  6. Problem 6New context

    A measurement variable X has variance 3. Find Var(3 X+7\displaystyle 3\, X + 7).

    Practise this exact problem →
  7. Problem 7New context

    A process population has standard deviation 8. For an independent random sample of size 4, what is the standard error of the sample mean?

    Practise this exact problem →
  8. Problem 8New context

    A binomial trial count X uses n=12\displaystyle n = 12 and p=12\displaystyle p = \frac{1}{2}. Find Var(X).

    Practise this exact problem →
Open stage USTAT 1.2 in the student workspace →

STAGE USTAT 1.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Estimation, Tests & Regression

Useful preparation: Expectation, Variance & the CLT

Open this topic in the university library →

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Inference connects observed data to a population or process through an explicit sampling model. An estimate identifies a plausible parameter value; an interval describes uncertainty from repeated sampling; a test measures incompatibility with a specified null model. A fitted regression line summarizes an association, while causal interpretation requires evidence about how the data were generated.

Confidence interval
A frequentist 95%\displaystyle 95\% interval procedure covers the fixed population parameter in 95%\displaystyle 95\% of repeated samples under its assumptions. The realized interval either contains that parameter or does not.
P-value
A p-value is the probability, assuming the null model and its assumptions, of a test statistic at least as incompatible with that model as the observed statistic, using the prespecified test direction.
Least-squares regression
A fitted line predicts the response from an explanatory variable by minimizing the sum of squared vertical residuals. A residual is observed response minus fitted response.

Conditions before calculation

  • The one-sample t interval is exact for independent observations from a normal population with unknown positive variance. Large-sample approximations require suitable conditions and can be poor with heavy tails or severe skewness.
  • Random sampling supports population generalization; random treatment assignment supports causal comparisons. A larger sample does not automatically remove selection bias or confounding.
  • Classical regression t inference assumes a correct linear conditional mean, independent errors, constant error variance, and normal errors for exact small-sample results. Fitting a line alone does not establish these assumptions.

Why the method works

1. Name the parameter and design

State the target population, parameter, and sampling process before selecting a formula. Distinguish the standard deviation of observations from the standard error of an estimator.

2. Match the reference distribution

When estimating a normal-population mean using sample standard deviation s, uncertainty in s leads to a t reference with n−1\displaystyle n - 1 degrees of freedom.

xˉ±t0.975,n−1sn,T=xˉ−μ0s/n\bar x\pm t_{0.975,n-1}\frac{s}{\sqrt n},\qquad T=\frac{\bar x-\mu_0}{s/\sqrt n}

3. Interpret limited evidence

A small p-value challenges the specified null model; it is not the probability that the null is true. Failure to reject does not prove equality. Report effect size and uncertainty alongside the decision.

Worked examples

EXAMPLE 1

An independent normal sample has n=16\displaystyle n = 16, mean 12, and sample standard deviation 4. Construct a 95%\displaystyle a\, 95\% mean interval and test mean 10 against a two-sided alternative.

  1. The estimated standard error is 1. Use t with 15 degrees of freedom and critical value approximately 2.131.

    SE⁡^=416=1\widehat{\operatorname{SE}}=\frac4{\sqrt{16}}=1

  2. Form the interval around the sample mean.

    12±2.131(1)=[9.869,14.131]12\pm2.131(1)=[9.869,14.131]

  3. The statistic for mean 10 is 2, below the two-sided cutoff in magnitude; 10 also lies inside the matching interval.

    T=12−101=2<2.131T=\frac{12-10}{1}=2<2.131

Conclusion: Do not reject mean 10 at the 5%\displaystyle 5\% level. This does not establish that the mean equals 10 or assign it a 95%\displaystyle a\, 95\% probability of lying in this realized interval.

EXAMPLE 2

Fit a least-squares line to points (0,1), (1,2), and (2,2).

  1. Compute the coordinate means, then centered sums.

    xˉ=1,yˉ=53,Sxx=2,Sxy=1\bar x=1,\quad\bar y=\frac53,\quad S_{xx}=2,\quad S_{xy}=1

  2. The slope divides the cross-product sum by the predictor squared-deviation sum.

    β^1=12,β^0=53−12=76\widehat\beta_1=\frac12,\qquad\widehat\beta_0=\frac53-\frac12=\frac76

  3. Check the residuals; they sum to zero because the model includes an intercept.

    e=(−16,13,−16),∑ei=0e=\left(-\frac16,\frac13,-\frac16\right),\quad\sum e_i=0

Conclusion: The fitted response is 76+x2\displaystyle \frac{7}{6} + \frac{x}{2}. These three observations support a descriptive fit, not a causal conclusion or verified error assumptions.

Pitfalls & counterexamples

Tempting claim: A p-value of 0.03 means there is a 3%\displaystyle a\, 3\% probability the null hypothesis is true.

Correction: The probability is calculated conditional on the null model; it is not a posterior probability for that model.

Test the claim: Different prior beliefs about a rare effect can yield different posterior probabilities even when a test gives the same p-value.

Check your understanding

Does a 95%\displaystyle a\, 95\% confidence interval contain 95%\displaystyle 95\% of individual observations?

Write a reason before opening the explanation. This reflection is not graded or saved.

Compare your reasoning

No.

A mean confidence interval targets the population mean. Prediction and tolerance intervals answer different questions about observations.

Connect this idea

Inference builds on sampling distributions and standard errors. Regression also connects to orthogonal projection; diagnostics and study design determine whether the fitted mathematics answers the scientific question.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts
  1. Problem 1

    A normal-mean interval uses estimate 50, known σ=8\displaystyle \sigma = 8, n=16\displaystyle n = 16 and multiplier z=2\displaystyle z = 2. Find its upper endpoint.

    Practise this exact problem →
  2. Problem 2

    A z test compares mean 22 to null mean 20, with known σ=6\displaystyle \sigma = 6 and n=9\displaystyle n = 9. Find the z statistic.

    Practise this exact problem →
  3. Problem 3

    A fitted line predicts y-hat=2 x+3\displaystyle 2\, x + 3. At x=4\displaystyle x = 4 the observed y is 14. Find the residual.

    Practise this exact problem →
  4. Problem 4

    A significance test uses level α=0.05\displaystyle \alpha = 0.05 and reports p=11000\displaystyle p = \frac{1}{1000}. Is p≤α\displaystyle p \le \alpha? Enter 1 for yes (reject at this level), 0 for no.

    Practise this exact problem →
  5. Problem 5New context

    A normally modeled sample mean interval uses estimate 51, known σ=8\displaystyle \sigma = 8, n=16\displaystyle n = 16 and multiplier z=2\displaystyle z = 2. Find its upper endpoint.

    Practise this exact problem →
  6. Problem 6New context

    A one-sample z test compares mean 23 to null mean 21, with known σ=6\displaystyle \sigma = 6 and n=9\displaystyle n = 9. Find the z statistic.

    Practise this exact problem →
  7. Problem 7New context

    A demand model predicts y-hat=2 x+4\displaystyle 2\, x + 4. At x=4\displaystyle x = 4 the observed y is 15. Find the residual.

    Practise this exact problem →
  8. Problem 8New context

    A pre-specified statistical test uses level α=0.05\displaystyle \alpha = 0.05 and reports p=21000\displaystyle p = \frac{2}{1000}. Is p≤α\displaystyle p \le \alpha? Enter 1 for yes (reject at this level), 0 for no.

    Practise this exact problem →
Open stage USTAT 1.3 in the student workspace →
Browse another level →
Numerive