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Worked Examples · Example 17
Q.

Calculate Z-Score for a normal distribution of length of 7 rare species of Indian butterfly that you have in your garden

Butterfly1234567
Length (in cm)2232516
Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★est
39% · 17/44 Questions
✓ Free question

With mean 33 cm and population SD 1.6901.690 cm, each length's Z-score is Z=x−31.690Z=\dfrac{x-3}{1.690}.

Z=x−xˉσZ=\dfrac{x-\bar{x}}{\sigma}, where xˉ=∑xn\bar{x}=\dfrac{\sum x}{n} and σ=∑(x−xˉ)2n\sigma=\sqrt{\dfrac{\sum (x-\bar{x})^{2}}{n}}.

  1. Mean. xˉ=2+2+3+2+5+1+67=217=3\bar{x}=\dfrac{2+2+3+2+5+1+6}{7}=\dfrac{21}{7}=3 cm.
  2. Squared deviations.
xxx−xˉx-\bar{x}(x−xˉ)2(x-\bar{x})^2
2-11
2-11
300
2-11
524
1-24
639
∑=20\sum=20
  1. Standard deviation. σ=207=2.857=1.690\sigma=\sqrt{\dfrac{20}{7}}=\sqrt{2.857}=1.690 cm.
  2. Z-scores Z=x−31.690Z=\dfrac{x-3}{1.690}:
ButterflyLengthZZ
12−0.592-0.592
22−0.592-0.592
3300
42−0.592-0.592
551.1831.183
61−1.183-1.183
761.7751.775
✓Final answer

xˉ=3\bar{x}=3 cm, σ=1.690\sigma=1.690 cm; Z-scores =−0.59, −0.59, 0, −0.59, 1.18, −1.18, 1.77=-0.59,\ -0.59,\ 0,\ -0.59,\ 1.18,\ -1.18,\ 1.77.

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