Skill: Calculate probabilities using normal distributions

Explanation and Free Practice Resources

A normal distribution models a continuous variable through a symmetric, bell-shaped density determined by its mean and standard deviation; probabilities are areas under the curve, found by standardizing values with z=(x−μ)/σz=(x-\mu)/\sigma and using a standard-normal table or technology. This includes calculating probabilities below, above, or between values and using complements, while recognizing that the probability of any exact single value is zero; multivariable and more advanced distribution theory are not included.

Detailed Explanation: Calculate probabilities using normal distributions

A normal distribution is a symmetric, bell-shaped model described by its:

  • Mean μ\mu: the center of the distribution
  • Standard deviation σ\sigma: how spread out the values are

To find a probability, first convert each value to a standard score:

z=x−μσz=\frac{x-\mu}{\sigma}

A zz-score tells you how many standard deviations a value is from the mean. Then use a standard-normal table or technology to find the area to the left of that zz-score.

Worked example

Suppose test scores are normally distributed with mean 7070 and standard deviation 88. Find the probability that a randomly selected score is between 6262 and 8282.

1. Write the probability

P(62<X<82)P(62<X<82)

2. Standardize both values

For 6262:

z=62−708=−1z=\frac{62-70}{8}=-1

For 8282:

z=82−708=1.5z=\frac{82-70}{8}=1.5

So the probability becomes:

P(−1<Z<1.5)P(-1<Z<1.5)

3. Use the standard-normal table

The table gives the area to the left of each zz-score:

P(Z<1.5)=0.9332P(Z<1.5)=0.9332 P(Z<−1)=0.1587P(Z<-1)=0.1587

4. Subtract the two areas

The area between the two values is:

P(−1<Z<1.5)=0.9332−0.1587=0.7745P(-1<Z<1.5)=0.9332-0.1587=0.7745

Therefore,

P(62<X<82)=0.7745\boxed{P(62<X<82)=0.7745}

There is a 77.45% probability that a randomly selected score is between 6262 and 8282.

For a value above a cutoff, use the complement:

P(X>a)=1−P(X<a)P(X>a)=1-P(X<a)

For a continuous normal variable, the probability of one exact value is zero:

P(X=70)=0P(X=70)=0

Probabilities refer to areas over intervals, such as being below, above, or between values.

Learn by doing: Calculate probabilities using normal distributions

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Statistics - Standard Deviation - Values and Z-Table Section to Z-Table Value


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