8.2a Z-Score Of A Normal Distribution

Posted on April 23, 2020 by user

8.2a z–Score of a Normal Distribution

Example:

(a) A normal distribution has a mean, µ = 50 and a standard deviation σ = 10. Calculate the standard score of the value X = 35.

(b) The masses of Form 5 students of a school are normally distributed with a mean of 60 kg and a standard deviation of 15 kg.

Find

(i) the standard score of the mass of 65 kg,

(ii) the mass of a student that corresponds to the standard score of – ½.

Solution:

(a)

X ~ N (µ, σ²).

X ~ N (50, 10²)

Z = \frac{X - μ}{σ} = \frac{35 - 50}{10} = - 1.5

(b)(i)

X – Mass of a Form 5 student

X ~ N (µ, σ²).

X ~ N (60, 15²)

Z = \frac{X - μ}{σ} = \frac{65 - 60}{15} = \frac{1}{3}

Hence, the standard score of the mass of 65 kg is ⅓.

(b)(ii)

Z = – ½,

\begin{array}{l} Z = \frac{X - μ}{σ} \\ - \frac{1}{2} = \frac{X - 60}{15} \\ X - 60 = - \frac{1}{2} (15) \\ X = 52.5 \end{array}

Hence, the mass of a Form 5 student that corresponds to the standard score of – ½ is 52.5 kg.