Normal Distribution Explained

Normal Distribution

• The normal distribution is a probability distribution.
• A normal distribution is symmetrical about the mean
• It forms a bell curve shape.
• More data points are centralized around the mean in a normal distribution.
• Normal distribution applies to continuous data.
• Standard normal distribution is centered around zero, representing the mean (µ).
• Three standard deviations above and below the mean typically cover the range.
• Sigma (σ) represents standard deviation, and variance is σ².

Properties of Normal Distribution Curves

• 99.8% of data falls within three standard deviations above and below the mean.
• 95% of data falls within two standard deviations above and below the mean.
• The area under the entire normal distribution curve equals one.
• Probabilities add up to one.
• To find the probability of a variable (x) being between two values, find the area under the curve between those values.
• The probability of x being exactly a specific number is infinitesimally small.
• The probability of x being less than the mean (µ) is 50% due to symmetry.

Notation

• "Is distributed" notation indicates a normal distribution.
• 'N' denotes normal distribution.
• "B" (in further maths) signifies binomial distribution.
• The normal distribution is defined by a mean (µ) and variance (σ²).
• x ~ N(µ, σ²) represents x is normally distributed with mean µ and variance σ².

Z-Value

• Z-value represents the number of standard deviations a value is above the mean.
• Each gap represents one standard deviation.
• Formula: Z = (x - µ) / σ.

Z Table

• Probabilities in the z-table are calculated as P(Z < z).
• The table provides the probability of being less than a given number of standard deviations
• If the table gives z is less than z, you must use this output from the table
• Symmetrical Curve: P(Z < -z) = 1 - P(Z < z)
• If the value you have is a negative, you must convert it by using the symmetry to transform it into a positive

Standardization

• Any distribution can be mapped onto the standard normal distribution using the formula: z = (x - µ) / σ
• x is the variable, µ is the mean, and σ is the standard deviation.
• This allows for finding z-values and probabilities using the z-table.

Probability Calculation for Flight Duration

• Calculation finds the probability of a flight from London to Malaga being less than 145 minutes
• Standardization converts the flight duration to a z-score for probability lookup
• The formula involves (variable - mean) / standard deviation: (145 - mean) / standard deviation
• Z-score calculation is (variable - mean) / standard deviation

Visual Representation and Reflection

• A diagram supports understanding, with zero representing the mean
• If the z-score is negative, draw picture first
• To find the area to the left of a negative z-score, reflect it to the positive side
• Because z-tables only give probabilities to the left, adjustment is needed for positive reflections
• The correct calculation is 1 - P(Z < 0.5) to find the area to the left of -0.5
• The probability of z being less than 0.5 has to be subtracted from 1, because it should not be higher than 0.5
• P(Z < 0.5) = 0.3085 is the probability of z being less than 0.5

Calculating Probabilities within a Range

• To find the probability within a range (e.g., 96 < X < 112), calculate z-scores for both bounds
• Process involves finding the probability of being less than the higher value and subtracting the probability of being less than the lower value
• Draw a picture to visualize the required area

Standardization with a Range

• Convert both values to z-scores using the standardization formula
• The probability = P(Z < 112 z-score) - P(Z < 96 z-score)
• Insert respective z-score calculations for each bound: P(Z < (112-100)/15) - P(Z < (96-100)/15)
• For P(Z < 0.8), the probability is 0.788.
• Calculations result in: P(Z < 0.8) - P(Z < -0.266)
• With negative z-score P(Z < -0.266), 1 - P(Z < 0.266) is required since the tables only provide positive lookups
• Approximation is necessary if the z-table has fewer decimal places

Final Calculation and Result

• The calculation converts the negative to a positive by reflecting it and figuring out its area will equal to the area to the right after changing it to a positive z value
• Rounding 0.266 to 0.27 provides a table-friendly value
• Rounded, get 1- P(Z < 0.27) = 0.6064.
• Final probability is: 0.788 - 0.6064 = 0.3945
• Final is: 0.3945

Further Examples

• Additional examples, for when it is not that easy to calculate the difference between 2 probabilities
• The probability of Z being less than 0.5 minus Z being less than 100 results in 0.1915
• Example involving range calculations result in a probability of 0.0627
• When you have a value being smaller and less than 0, it has requires further calculations

Reverse Calculations

• Starting with a known probability and working backwards to find the corresponding z-score value
• This finds z-value which has the probability of z being less than little z, and it equals a value of 0.87
• Consult the standard normal distribution table and find the z value corresponding to a probability of 0.87
• Use the Z-table value closest to probability 0.87 in distribution
• Closest z value to 0.87: 1.13
• The value of z if the probability is greater than z value is 0.23: 0.74

Probabilities Less Than 0.5

• This looks for z when P(Z < z) = 0.1
• Since 0.1 is less than 0.5, the z-score will be negative because 50% is mean/median
• Using symmetry, the probability of z being less than z is the same as being to the right now,
• Z = -1.29

In Summary

• If probability is "less than" and over 0.5, look up z-table value
• For "greater than", adjustments are sometimes needed to use z table for "less than" lookup, or convert by flipping it
• If "less than" and probability is less than 0.5, value for z is negative

Drawing a diagram

• The area to the right must be greater than 0.86: this number must be negative
• Multiply the graph by negative 1, to move the position of A to the negative range and it makes it a less than graph
• With a drawn and updated diagram, you can then read a table to find its value
• Thus, the probability of Z being greater than A, is the same with Z being smaller than the negative number, multiplied by negative 1

Working With Examples

• When probability is being "less than", everything is ok, you can look at the table
• When it involves minuses, further calculations and consideration is needed
• For E, the answer is 1.65

Z-Score Formula

• Introduces the Z-score formula: Z = (X - Mean) / Standard Deviation
• X is 130
• Mean is 100
• Standard Deviation is 15
• In this case, 100-130/15 = 2### Calculating Probability Above or Below a Value
• To find the probability of an IQ being above 115, calculate P(X > 115).
• Convert X to Z using the formula: Z = (X - μ) / σ, where μ = 100 and σ = 15
• P(X > 115) translates to P(Z > (115 - 100) / 15) = P(Z > 1)
• Since tables usually provide P(Z < z), use the complement rule: P(Z > 1) = 1 - P(Z < 1)
• Look up Z = 1 in the Z-table to find P(Z < 1) = 0.8413.
• Calculate the final probability: 1 - 0.8413 = 0.1587.

Calculating Z-Value from a Probability

• To find the Z-value corresponding to P(Z < z) = 0.58, look for 0.58 in the Z-table.
• Find the closest value to 0.58 in the table, which is 0.5793, corresponding to Z = 0.20.
• For P(Z > z) = 0.22, use the complement rule: 1 - P(Z < z) = 0.22, so P(Z < z) = 0.78
• Look for 0.78 in the table, identify the closest value as 0.7794, and find the corresponding Z.
• For P(Z < z) = 0.15, consider the symmetry of the normal distribution; find P(Z < z) = 0.85 first, which is 1.04, then take the negative.
• For P(Z > z) = 0.95, find P(Z < z) = 0.05, and recognize it will be a negative Z-value due to its location of being less than 0.5.
• The Z-value will be negative and can be found using symmetry, use value of 0.95 (the inverse of 0.05)

Retrieving Original Values

• To find the IQ representing the bottom 78% of the population, state the problem: P(X < x) = 0.78.
• Standardize to P(Z < z) = 0.78, find the Z-value in the table (approximately 0.77).
• Reverse the Z-formula: X = Z * σ + μ = 0.77 * 15 + 100 = 111.55
• For the bottom 90%, P(X < x) = 0.9, standardize to P(Z < z) = 0.9.
• Find Z = 1.28 in the table, then X = 1.28 * 15 + 100 = 119.2.
• For the bottom 30%, P(X < x) = 0.3, which means P(Z < z) = 0.3.
• Since 0.3 < 0.5, use symmetry: find Z for 0.7 (1 - 0.3), then negate it to get Z = -0.52.
• Then X = -0.52 * 15 + 100 = 92.2 or 92.14
• To find the IQ that 80% of the population has more than, use P(X > x) = 0.8.
• Convert to P(Z > z) = 0.8, which is 1 - P(Z < z) = 0.8, so P(Z < z) = 0.2.
• Use symmetry to find the negative Z-value: Z = -0.84 (from the table for 0.8), then X = -0.84 * 15 + 100 = 87.4.

Exam-Style Questions

• Cement Bag Weight Example:
  • X ~ N(50, 2^2) where the mean is 50 kg and standard deviation is 2 kg.
  • Find the weight exceeded by 99% of bags: P(X > x) = 0.99.
  • Find equivalent P(Z>z) = 0.99, covert to P(Z<z) = 0.01 (which also cannot be found)
  • Use the probability of Z being less than a negative Z P(Z<-z) = 0.99, find P(Z>z) = 0.01
  • Recognize the use of the symmetry to change it to a negative value, then P(Z < -2.32), so finding for 0.99 which is 2.32 using the symmetry rules
  • Z = -2.32 = (X - 50)/2 so X = -2.32 * 2 + 50 which equals = 45.36.
• Calculator use:
  • On calculators such as the Casio Classwiz: Mode -> 7 (Distribution) -> 3 (Inverse Normal).
  • Input area, standard deviation, and mean, find value of X
• 100m Run Time Example:
  • X ~ N(16.12, 1.6^2), find the time for the fastest 30% of children.
  • Find P(X > x) = 0.30, convert to 0.7.
  • Due to properties, this reads as P(z<-0.52) = 0.7 using symmetry.
  • So we can use that, or 0.3, or 0.7 value
  • Therefore Z = -0.52 = (X - 16.12) / 1.6 -> X = 15.28. Confirm with calculator.
  • The most important step is to draw the normal distribution diagram, if necessary.

Harder Reverse Probability Questions

• Symmetric Probability Range Example:
  • find "a" such that P(-a < Z < a) = 0.7, draw distribution diagram.
  • Split probability in half: 0.7/2 = 0.35 each side from mean.
  • Find P(Z < a) such that area represents 0.5 (below mean) + 0.35 to 0.85: P(Z<a) = 0.85.
  • Look up corresponding Z / a value in table for P(Z < a) = 0.85 equals 1.04
• Normal Distribution Inequality Example:
  • X ~ N(30, 5^2); find "k" where P(30 < X < k) = 0.2, noting k > 30.
  • Given 30 is at mean, find P(X < k) which translates to P(Z < z) = 0.5 + 0.2 = 0.7.
  • Find equivalent z-score where P(Z < z) = 0.7, look up 0.7 equals 0.52.
  • So 0.52 = (k - 30) / 5. Invert: k = 32.6.

Quartiles and Percentiles

• Lower Quartile:

  • Given μ = 100 and σ = 15, find the value of the divide for probability of X being less than lower quartile is 0.25. i.e z = 0.25
  • Translate to P(Z < z) = 0.25
  • But it doesn't work as this reads as. 0.75 = z which flips the z valueto the negative side
  • Find z equals -0.64 (roughly this value) Then X = -0.67 * 15 + 100 = 89.95 = lower quartile

• Upper Quartile

  • Upper quartile for x < 75 (0.75) translate this, find that the Upper quartile value will have a positive z value or 0.67 (z)
  • Then the upper quartile would read as X-100/15 i.e the 0.67*15 +100 = 110.05 (roughly)

• Important note, you can minus from the mean to find upper and lower quartiles

• 70th Percentile

  • Read like a normal distribution. X< 70th Percentile, read P<0.7
  • So z we know = to equal 0.7 which means = roughly 0.52 (check table)
  • P70 - 100/ 15 = x translates to roughly 107.6

Testing Understanding

• Heights example
  • X is distributed as normal with a mean of 162 and standard deviation of 7.5,
  • Sergio, who is at the 60th Percentile. Find that percentage
  • SO 0.6 of x/z = standard deviation , z is found to = 0.25
  • So, x =62/7.5 = z
  • i.163.16-162/ 7.5/
• Distances to work
  • d distributed with a mean of 30 standard deviation of 8KM
  • So,
  • Find probability of someone traveling more than 20km (20KM Z) = translate Z and apply to distribution

Mean and Standard Deviation Questions

• Example 1, mean unknown, standard deviation equals three
• Probability of X being bigger than 20 equals 0.2, find
• Find reverse standard distribution,

Normal Distribution Problem Solving

• When dealing with a normal distribution with an unknown standard deviation, the Z-score formula and probability concepts are critical.

Finding Standard Deviation

• Probability of X < 46 is 0.2119, which is less than 0.5, indicating a need to find the equivalent probability on the other side of the mean
• Finding the negative Z-score: Probability of Z < -Z = 1 - 0.2119 = 0.7881.
• The Z-score corresponding to 0.7881 is 0.8, so -Z = -0.8.
• Applying the Z-score formula: (46 - 50) / standard deviation = -0.8.
• Standard deviation is calculated as (46 - 50) / -0.8 = 5.
• A negative standard deviation indicates an error in the calculation.

Solving for Mean and Standard Deviation with Two Unknowns

• When two unknowns (mean and standard deviation) exist, setting up a system of equations is typically needed.
• Two equations is set up based on given probabilities.
• Equation 1: Probability of X > 35 = 0.025.
• Probability of X < 35 = 1 - 0.025 = 0.975.
• The Z-score corresponding to 0.975 is 1.96.
• First equation: (35 - mean) / standard deviation = 1.96, which can be rearranged to 35 - mean = 1.96 * standard deviation.
• Equation 2: Probability of X < 15 = 0.1469, which is less than 0.5, implying a need to use the negative Z-score.
• Finding the negative Z-score: Probability of Z < -Z = 1 - 0.1469 = 0.8531.
• The Z-score is -1.05.
• Second equation: (15 - mean) / standard deviation = -1.05, which is rearranged to 15 - mean = -1.05 * standard deviation.

Solving the Equations

• Subtracting the two equations to eliminate the mean.
• Calculation: 20 = 1.96 * standard deviation + 1.05 * standard deviation = 3.01 * standard deviation.
• Find standard deviation: standard deviation = 20 / 3.01 ≈ 6.64 (2 decimal places).
• Substituting the standard deviation value back into one of the equations to find the mean. Mean ≈ 22.0 (to one decimal place) after calculations and substitutions.