Question

The average length of a sample of 50 leaves measured by a botanist was 40 mm with a standard deviation of 21 mm. Assuming normal deviation find the limits which have a 95% chance of including the expected length of the leaf.
1. 42.35 - 37.65
2. 34.12 - 45.88
3. 43.12 - 36.88
4. 44.58 - 35.42

Answer 1

Correct Answer - Option 2 : 34.12 - 45.88

Given

Sample size = 50 mm

Mean = x̅ = 40 mm

Standard deviation = σ = 21 mm

Concept

For the population means μ (or expected length; the standard is normal and it is given by

x̅ ± 1.96 σ/√n

Calculation

40 ± (1.96 × 21)/√50

⇒ 40 ± 5.88

⇒ 34.12 or 45.88

∴ The limits which have a 95% chance of including the expected length of the leaf was  34.12 or 45.88