Finding an unknown mean or standard deviation

To find an unknown mean or standard deviation, turn the given probability into a z-value with the table, then solve z = (x − μ) ÷ σ for what you do not know. A sketch of the curve first keeps the sign of z right.

This lesson builds on standardising a normal random variable and reading tail probabilities in the probability distributions chapter.

Worked example: unknown σ

X ~ N(40, σ²) and P(X > 50) = 0.0668. Find σ.

The upper-tail probability is 0.0668, so the table gives z = 1.5. The value 50 is above the mean, which agrees with a positive z.

1. Write (50 − 40) ÷ σ = 1.5.
2. Solve: σ = 10 ÷ 1.5 = 6.67 (3 s.f.).

Worked example: unknown μ

X ~ N(μ, 5²) and P(X < 32) = 0.1587. Find μ.

The lower-tail probability is below 0.5, so 32 is below the mean and z is negative. The table gives z = −1.

1. Write (32 − μ) ÷ 5 = −1.
2. Solve: 32 − μ = −5, so μ = 37.

Worked example: both unknown

X ~ N(μ, σ²), with P(X > 30) = 0.1587 and P(X < 18) = 0.0228.

The first probability gives z = 1 and the second gives z = −2 (from the lower tail). So:

• (30 − μ) ÷ σ = 1, which gives 30 − μ = σ.
• (18 − μ) ÷ σ = −2, which gives 18 − μ = −2σ.

Subtract the second from the first: 12 = 3σ, so σ = 4. Then μ = 30 − 4 = 26.

Check: (30 − 26) ÷ 4 = 1 and (18 − 26) ÷ 4 = −2. Both match.

The mistake that costs marks

The common slip is to use a lower-tail probability as if it were an upper-tail one. Given P(X < 32) = 0.1587, a student reads z = 1 and writes (32 − μ) ÷ 5 = 1, getting μ = 27.

The check: 32 would then be above the mean, which makes P(X < 32) larger than 0.5, not 0.1587. A quick sketch of the curve catches the sign.

Check yourself

X ~ N(μ, 3²) and P(X > 25) = 0.3085. Find μ.

What to study next

Test the whole chapter with the probability distributions practice set. The word-problem structure worksheet helps separate each probability statement into its own equation.

If the sign of z keeps going wrong, see online one-to-one Additional Mathematics tuition.