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Normal Distribution
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The normal (bell-shaped) distribution \(X\sim\mathrm{N}(\mu,\sigma^2)\), its symmetry about \(\mu\), z-values \(z=\frac{x-\mu}{\sigma}\) and standardization to \(Z\sim\mathrm{N}(0,1)\), the empirical rule, and calculator-based normal and inverse normal probability calculations.
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The Normal Distribution
SL 4.9
The normal distribution, often called the bell curve, is a symmetric, bell-shaped probability distribution widely used to model natural variability and measurement errors. It appears frequently in natural settings because averaging many small, independent effects tends to produce results that cluster around a central value, naturally forming a bell-shaped distribution.
The normal distribution is characterized by its mean, μ, and standard deviation, σ, which completely and uniquely describe both the central value and how "spread out" the curve is. By convention, we describing the normal distribution by writing X∼N(μ,σ2). Notice that σ2 is the variance, not the standard deviation.
The probability that X is less than a given value a, written P(X<a), is equal to the area under the curve to the left of x=a:
It follows that the total area under the curve is 1, which is required as the probabilities must sum to 1.
The Bell Curve properties
SL 4.9
Because of the symmetry of the normal distribution, we know that
P(X>μ)=P(X<μ)=21=0.5🚫
Further, for any real number a,
P(X>μ+a)=P(X<μ−a)
Try visualizing the relationship between percentage of data points and percentage of area by changing the bounds of highlighted area with the tool below:
In practice, you'll find probabilities of normally distributed random variables using your calculator.
The Empirical Rule
SL 4.9
It is also useful (but not often required) to know the empirical rule:
P(μ−σ<X<μ+σ)P(μ−2σ<X<μ+2σ)P(μ−3σ<X<μ+3σ)≈68%≈95%≈99.7%
Normal calculations
SL 4.9
To calculate P(a<X<b) for X∼N(μ,σ2) on your GDC, press 2nd →distr (on top of vars) to open the probability distribution menu. Select normalcdf( with your cursor. Type the value of a after "lower," the value of b after "upper," the value of μ after "μ," and the value of √σ2=σ after σ (since the calculator asks for standard deviation, not variance). Then click enter twice and the calculator will return the value of P(a<X<b).
If you want to find a one-sided probability like P(a<X), enter the value ±1×1099 as the upper or lower bound.
Under the hood, the calculator is finding the area under the normal curve between x=a and x=b:
Inverse Normal Calculations
SL 4.9
The calculator can also perform inverse normal calculations. That is, given the mean μ, the standard deviation σ, and the probability P(X<a)=k, the calculator can find the value a.
On your GDC, press 2nd →distr (on top of vars) to open the probability distribution menu. Select invNorm( with your cursor. Type the value of k after "area," the value of μ after "μ," and the value of √σ2=σ after σ (since the calculator asks for standard deviation, not variance). Then click enter twice and the calculator will return the value of a.
Note the calculator specifically returns the value of the "left end" of the tail. To find the value of some b when given P(b<X)=k, enter the value of 1−k (the complement) as the area.
Z-values
SL 4.12
For a random variable X with X∼N(μ,σ2) and a specific value x in the probability distribution, the z-value of x is defined as
z=σx−μ📖
The z-value measures how many standard deviations a value x lies above or below the mean. Put differently,
Z=σX−μ⟹Z∼N(0,1)
The random variable follows a normal distribution X∼N(μ,72). Given that P(X<60)=0.8, find μ.
On the standard normal distribution N(0,1), the area is 0.8 when z=invNorm(0.8, 0, 1) =0.84162. So
z=760−μ=0.84162⇒μ=54.1
Example: The random variable X follows a normal probability distribution X∼N(11,σ2). Given that P(22<X)=0.4, find σ.
If P(22<X)=0.4 then P(X<22)=1−0.4=0.6.
On the standard normal distribution N(0,1), the area is 0.6 when z=invNorm(0.6, 0, 1) =0.253347. So
z=σ22−11=0.253347⇒σ=0.25334711=43.4
Let X be a random variable with X∼N(μ,σ2). Given that P(X>2)=0.1 and P(X<0.5)=0.05, find μ and σ.
If P(X>2)=0.1 then P(X<2)=1−0.1=0.9. On the standard normal distribution Z∼N(0,1), the area is 0.9 when z1=invNorm(0.9,0,1) =1.28. This gives the equation
1.28=σ2−μ(1)
We also know that P(X<0.5)=0.05. On the standard normal distribution Z∼N(0,1), the area is 0.05 when z2=invNorm(0.05,0,1) =−1.64. This gives the equation
−1.64=σ0.5−μ(2)
Solving equations (1) and (2),
1.28=σ2−μ⟹μ=2−1.28σ
−1.64=σ0.5−μ⟹−1.64=σ0.5−(2−1.28σ)⟹σ=0.514
1.28=0.5142−μ⟹μ=1.34
So σ=0.514, μ=1.34.
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