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๐ŸŽฏโญ INTERACTIVE LESSON

Continuous Random Variables

Learn step-by-step with interactive practice!

Continuous Random Variables - Complete Interactive Lesson

Part 1: Introduction to Probability Distributions

๐Ÿ“Š Introduction to Probability Distributions

Part 1 of 7 โ€” The Normal Model


The Normal Distribution

The most important continuous probability distribution in statistics.

Key Properties:

  • Bell-shaped and symmetric
  • Mean = median (at the center)
  • Described completely by ฮผ\mu (mean) and ฯƒ\sigma (standard deviation)
  • Notation: XโˆผN(ฮผ,ฯƒ)X \sim N(\mu, \sigma)

The 68-95-99.7 Rule (Empirical Rule)

RangePercent of Data
ฮผยฑ1ฯƒ\mu \pm 1\sigmaโ‰ˆ68%
ฮผยฑ2ฯƒ\mu \pm 2\sigmaโ‰ˆ95%
ฮผยฑ3ฯƒ\mu \pm 3\sigmaโ‰ˆ99.7%

Example

Heights of adult women: N(64,2.5)N(64, 2.5) inches

  • 68% are between 61.5 and 66.5 inches
  • 95% are between 59 and 69 inches
  • 99.7% are between 56.5 and 71.5 inches

Concept Check ๐ŸŽฏ

Empirical Rule Practice ๐Ÿงฎ

SAT scores follow N(1060,195)N(1060, 195).

1) 68% of scores fall between ___ and ___ (give the lower bound)

2) Whatโ€™s the upper bound?

3) What percent of scores fall below 10601060?

Part 2: Normal Distribution Properties

๐Ÿ“ Z-Scores

Part 2 of 7 โ€” Standardizing Values


The Z-Score Formula

z=xโˆ’ฮผฯƒz = \frac{x - \mu}{\sigma}

A z-score tells you how many standard deviations a value is from the mean.

Z-ScoreInterpretation
z=0z = 0At the mean
z=1z = 11 SD above the mean
z=โˆ’2z = -22 SD below the mean

Standard Normal Distribution

When we standardize, we convert to ZโˆผN(0,1)Z \sim N(0, 1).

Example

Test scores: ฮผ=75\mu = 75, ฯƒ=8\sigma = 8. A student scores 91.

z=91โˆ’758=168=2.0z = \frac{91 - 75}{8} = \frac{16}{8} = 2.0

The student scored 2 standard deviations above the mean.

Concept Check ๐ŸŽฏ

Z-Score Calculations ๐Ÿงฎ

Heights: ฮผ=170\mu = 170 cm, ฯƒ=6\sigma = 6 cm.

1) Z-score for a height of 182 cm?

2) Z-score for a height of 158 cm?

3) A z-score of 0 corresponds to what height? (in cm)

Part 3: Standard Normal Distribution

๐Ÿ“– Normal Probabilities

Part 3 of 7 โ€” Using the Standard Normal Table


Finding Probabilities

The Standard Normal Table (Table A) gives P(Z<z)P(Z < z) โ€” the area to the LEFT of zz.

Three Cases

WantFormula
P(Z<z)P(Z < z)Read directly from table
P(Z>z)P(Z > z)1โˆ’P(Z<z)1 - P(Z < z)
P(a<Z<b)P(a < Z < b)P(Z<b)โˆ’P(Z<a)P(Z < b) - P(Z < a)

Example

P(Z<1.25)P(Z < 1.25): Look up z=1.25z = 1.25 in Table A โ†’ 0.89440.8944

P(Z>1.25)=1โˆ’0.8944=0.1056P(Z > 1.25) = 1 - 0.8944 = 0.1056

P(โˆ’1<Z<1.25)=P(Z<1.25)โˆ’P(Z<โˆ’1)=0.8944โˆ’0.1587=0.7357P(-1 < Z < 1.25) = P(Z < 1.25) - P(Z < -1) = 0.8944 - 0.1587 = 0.7357

Working Backwards (Inverse Normal)

To find the z-score for a given percentile:

  • 90th percentile: z=1.28z = 1.28
  • 95th percentile: z=1.645z = 1.645
  • 97.5th percentile: z=1.96z = 1.96

Concept Check ๐ŸŽฏ

Normal Probability ๐Ÿงฎ

P(Z<1.0)=0.8413P(Z < 1.0) = 0.8413 and P(Z<โˆ’0.5)=0.3085P(Z < -0.5) = 0.3085.

1) P(Z>1.0)=?P(Z > 1.0) = ?

2) P(โˆ’0.5<Z<1.0)=?P(-0.5 < Z < 1.0) = ?

3) What percent of a Normal distribution falls between z=โˆ’0.5z = -0.5 and z=1.0z = 1.0?

Part 4: Z-Scores and Applications

๐Ÿ” Assessing Normality

Part 4 of 7 โ€” Is the Data Normal?


Look at a Graph of the Data

GraphConsistent with a Normal modelNot consistent
Histogram or dotplotOne peak, roughly symmetricStrong skew, two peaks, gaps
BoxplotMedian near the middle of the box, similar whiskers, no outliersLong whisker on one side, outliers
68-95-99.7 RulePercents within 1, 2, 3 SDs near 68%, 95%, 99.7%Percents far from those values

๐Ÿ”‘ Real data are never exactly Normal. The question is whether a Normal model is a reasonable approximation.


When Can We Use Normal Methods?

  • Data: a Normal model fits the data only if the graph looks roughly symmetric and unimodal with no outliers.
  • Sample means: the sampling distribution of xห‰\bar{x} is approximately Normal if the population is Normal or nโ‰ฅ30n \geq 30 (Central Limit Theorem).
  • Small samples: check a graph of the sample for strong skewness and outliers.

Concept Check ๐ŸŽฏ

Normality Assessment ๐Ÿงฎ

1) A histogram of the data shows two distinct peaks. Is a Normal model reasonable? (yes/no)

2) For n=50n = 50 observations from a skewed population, is the sampling distribution of xห‰\bar{x} approximately Normal? (yes/no)

3) What rule compares the percent of data within 1, 2 and 3 SDs to a Normal model? (state the name)

Part 5: Sampling Distributions

๐Ÿ“Š Sampling Distributions

Part 5 of 7 โ€” Sampling Distributions of xห‰\bar{x} and p^\hat{p}


Sampling Distribution of xห‰\bar{x}

If we take many samples of size nn from a population with mean ฮผ\mu and SD ฯƒ\sigma:

xห‰โˆผN(ฮผ,ฯƒn)\bar{x} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)

PropertyValue
Meanฮผxห‰=ฮผ\mu_{\bar{x}} = \mu
Standard deviationฯƒxห‰=ฯƒn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
ShapeNormal (if population Normal OR nโ‰ฅ30n \geq 30)

Sampling Distribution of p^\hat{p}

For proportions from samples of size nn where population proportion is pp:

p^โˆผN(p,p(1โˆ’p)n)\hat{p} \sim N\left(p, \sqrt{\frac{p(1-p)}{n}}\right)

Conditions: npโ‰ฅ10np \geq 10 and n(1โˆ’p)โ‰ฅ10n(1-p) \geq 10

Key Insight

As nn increases, the standard deviation decreases โ€” larger samples give more precise estimates.

Concept Check ๐ŸŽฏ

Sampling Distribution ๐Ÿงฎ

Population: ฮผ=500\mu = 500, ฯƒ=100\sigma = 100. Sample size n=25n = 25.

1) Mean of the sampling distribution of xห‰\bar{x}?

2) Standard deviation of xห‰\bar{x}?

3) P(xห‰>520)=P(Z>?)P(\bar{x} > 520) = P(Z > ?) โ€” what is the z-score?

Part 6: Problem-Solving Workshop

๐Ÿ† Problem-Solving Workshop

Part 6 of 7 โ€” AP-Style Practice


Common AP Problem Types

  1. Find a probability using z-scores and Table A
  2. Find a percentile (inverse Normal)
  3. Compare values from different distributions using z-scores
  4. Sampling distribution questions about xห‰\bar{x} or p^\hat{p}

Comparing Across Distributions

Who performed better?

  • Student A: scored 680 on SAT (ฮผ=500,ฯƒ=100)(\mu = 500, \sigma = 100)
  • Student B: scored 28 on ACT (ฮผ=21,ฯƒ=5)(\mu = 21, \sigma = 5)

zA=680โˆ’500100=1.80zB=28โˆ’215=1.40z_A = \frac{680-500}{100} = 1.80 \qquad z_B = \frac{28-21}{5} = 1.40

Student A has a higher z-score, so Student A did relatively better.

Concept Check ๐ŸŽฏ

Comparing Distributions ๐Ÿงฎ

Math: ฮผ=72,ฯƒ=10\mu = 72, \sigma = 10. English: ฮผ=80,ฯƒ=5\mu = 80, \sigma = 5. A student scores 87 in Math and 88 in English.

1) Z-score in Math?

2) Z-score in English?

3) In which class did the student do relatively better? (math/english)

Part 7: Mixed Review

๐Ÿ“ Mixed Review

Part 7 of 7 โ€” Comprehensive Review


Key Formulas

FormulaPurpose
z=(xโˆ’ฮผ)/ฯƒz = (x - \mu)/\sigmaStandardize a value
x=ฮผ+zฯƒx = \mu + z\sigmaUnstandardize (find value from z)
ฯƒxห‰=ฯƒ/n\sigma_{\bar{x}} = \sigma/\sqrt{n}Standard error of the mean
ฯƒp^=p(1โˆ’p)/n\sigma_{\hat{p}} = \sqrt{p(1-p)/n}Standard error of a proportion

Checklist

  • Normal distribution: bell-shaped, symmetric, ฮผ\mu and ฯƒ\sigma
  • 68-95-99.7 Rule
  • Z-scores and Table A
  • Sampling distributions of xห‰\bar{x} and p^\hat{p}
  • Checking normality with graphs and the 68-95-99.7 rule
  • Central Limit Theorem

Concept Check ๐ŸŽฏ

Final Challenge ๐Ÿงฎ

XโˆผN(200,25)X \sim N(200, 25). A sample of n=100n = 100 is taken.

1) ฯƒxห‰=?\sigma_{\bar{x}} = ?

2) P(xห‰>205)P(\bar{x} > 205): first find z=?z = ?

3) Using Table A, P(Z>2)โ‰ˆ?P(Z > 2) \approx ?