How to find probability with mean and standard deviation

What is the z-score method for finding probability?

The z-score method transforms any normal distribution into the standard normal distribution, which has a mean of 0 and a standard deviation of 1. This transformation allows you to use a single reference table to find probabilities for any normally distributed dataset.

The z-score formula is:

$z=\frac{x-\mu }{\sigma }$

Where:

• $x$ = the value of interest
• $\mu$ = the population mean
  
• $\sigma$ = the population standard deviation

For sample data where population parameters are unknown, use the sample standard deviation ($s$) instead: 

$z=\frac{x-\overline{x}}{s}$

Where:

• $\bar{x}$ = the sample mean
• $s$ = the sample standard deviation

Use population standard deviation ($\sigma$) when you have data for an entire population or when population parameters are known. Use sample standard deviation ($s$) when working with a subset of data drawn from a larger population.

How do you read a standard normal table?

A standard normal table (z-table) displays cumulative probabilities for the standard normal distribution. The table provides $P(Z \leq z)$, which represents the area under the curve to the left of a given z-score.
$P,(Z\le z)$

Table structure

The leftmost column lists z-scores to one decimal place (the ones and tenths digits). The top row lists the hundredths digit. The intersection of a row and column gives the cumulative probability.

Reading process

1. Locate the row corresponding to the first two digits of your z-score (ones and tenths place)
2. Locate the column corresponding to the hundredths digit
3. Read the probability value at the intersection

For z = 1.23:

• Find row 1.2 in the left column
• Find column 0.03 in the top row
• Read the value 0.8907 at the intersection

This value means $P(Z \leq 1.23) = 0.8907$, or 89.07% of values fall below a z-score of 1.23.

Negative z-scores

Negative z-scores indicate values below the mean. Z-tables for negative values work identically. For z = -1.23, find row -1.2 and column 0.03 to get 0.1093, meaning 10.93% of values fall below this point.

How do you find probability below a value?

Finding the probability below a value (left-tail probability) uses the z-table directly without adjustment.

Steps

1. Calculate the z-score using the formula $z = \frac{x - \mu}{\sigma}$
2. Locate the z-score in the standard normal table
3. Read the table value, which equals $P(X < x)$
   

Application in test scores

A standardized test has a mean score of 500 and a standard deviation of 100. To find the probability that a randomly selected student scores below 620:

, which represents the

a

rea under the curve to the left of a given z-score. , which represents the area under the curve to the left of a given z-score. 

$z=\frac{620-500}{100}=\frac{120}{100}=1.20$

Looking up z = 1.20 in the table gives 0.8849. The probability that a student scores below 620 is 88.49%.

How do you find probability above a value?

Finding the probability above a value (right-tail probability) requires subtracting the table value from 1, since z-tables provide left-tail (cumulative) probabilities.

Steps

1. Calculate the z-score using $z = \frac{x - \mu}{\sigma}$
2. Look up the z-score in the table to find $P(Z \leq z)$
   
3. Subtract from 1: $P(X>x)=1-P(Z\le z)$

Application in quality control

A manufacturing process produces bolts with a mean diameter of 10 mm and a standard deviation of 0.2 mm. To find the probability that a bolt exceeds 10.5 mm:

$z=\frac{10.5-10}{0.2}=\frac{0.5}{0.2}=2.50$

The table value for z = 2.50 is 0.9938.

$P(X>10.5)=1-0.9938=0.0062$

The probability that a bolt exceeds 10.5 mm diameter is 0.62%.

How do you find probability between two values?

Finding the probability between two values requires calculating two z-scores and subtracting their corresponding probabilities.

Steps

1. Calculate z-scores for both values: $z_1 = \frac{x_1 - \mu}{\sigma}$ and $z_2 = \frac{x_2 - \mu}{\sigma}$
2. Look up both z-scores in the table
3. Subtract the smaller probability from the larger: $P(x_1<X<x_2)=P(Z\le z_2)-P(Z\le z_1)$

Application in manufacturing tolerances

A machine produces components with a target length of 50 cm, a mean of 50 cm, and a standard deviation of 0.5 cm. Acceptable tolerance is 49 cm to 51 cm. To find the probability a component falls within tolerance:

$z_1=\frac{49-50}{0.5}=\frac{-1}{0.5}=-2.00$

$z_2=\frac{51-50}{0.5}=\frac{1}{0.5}=2.00$

From the z-table:

• $P(Z \leq 2.00) = 0.9772$
  
• $P(Z \leq -2.00) = 0.0228$
  

$P(49<X<51)=0.9772-0.0228=0.9544$

The probability that a component falls within the acceptable tolerance range is 95.44%.

What is the empirical rule for quick probability estimation?

The empirical rule (68-95-99.7 rule) provides quick probability estimates for normal distributions without requiring z-table lookups. This rule states that for any normal distribution:

• 68% of data falls within 1 standard deviation of the mean
• 95% of data falls within 2 standard deviations of the mean
• 99.7% of data falls within 3 standard deviations of the mean

Mathematical representation

$P(\mu -1\sigma <X<\mu +1\sigma )\approx 0.68$

$P(\mu -2\sigma <X<\mu +2\sigma )\approx 0.95$

$P(\mu -3\sigma <X<\mu +3\sigma )\approx 0.997$

Using the empirical rule

For a test with mean 100 and standard deviation 15:

• 68% of scores fall between 85 and 115
• 95% of scores fall between 70 and 130
• 99.7% of scores fall between 55 and 145

The empirical rule works for quick estimates when values align with whole standard deviations from the mean. Use z-scores and tables for precise probabilities or values that do not fall exactly at 1, 2, or 3 standard deviations.

What are the limitations of this method?

The z-score method yields exact probabilities only for normally distributed data. Several conditions and alternatives apply when data does not meet this assumption.

Normality requirement

Z-tables assume the underlying data follows a normal distribution. Verify normality through histograms, Q-Q plots, or statistical tests before applying this method.

Chebyshev's inequality for non-normal data

For distributions of unknown shape, Chebyshev's inequality provides minimum probability bounds:

$P(|X-\mu |<k\sigma )\ge 1-\frac{1}{k^2}$

This guarantees at least 75% of data within 2 standard deviations and 89% within 3 standard deviations for any distribution, though these bounds are less precise than normal distribution calculations.

Normal approximation conditions

The normal distribution can approximate other distributions under certain conditions. For binomial distributions, the normal approximation applies when:

$np\ge 5\text{and}n(1-p)\ge 5$

Where $n$ is the number of trials and $p$ is the probability of success.

Sample size considerations

When using sample statistics instead of population parameters, the t-distribution provides more accurate probabilities for small samples (typically n < 30). The t-distribution accounts for additional uncertainty introduced by estimating population parameters from sample data.