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Goodness of fit: normal and continuous uniform models

Edexcel IAL Mathematics · WST03 Statistics 3

Continuous data come grouped into classes. The expected frequency of each class is N times the model probability of that class; the specification says long calculations will not be set, so questions often give most of the expected values.

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Continuous uniform model

For U[a, b], the probability of a class is (class width)/(b - a). So expected frequencies are in proportion to class width.

Arrivals over 60 minutes, N = 120, classes of width 10, 10, 20, 20: E = 20, 20, 40, 40.

If a and b are both given, ν = cells - 1.

Minutes after the hour0 to 1010 to 2020 to 4040 to 60
Arrivals26153742

120 arrival times

Normal model

For N(μ, σ2) find each class probability by standardising, with the end classes open (h < 160 and h ≥ 180) so the probabilities total 1.

Heights N(170, 82), N = 200: P(h < 160) = Φ(-1.25) = 0.1056, so E = 21.13; P(160 ≤ h < 170) = 0.3944, E = 78.87; the other two by symmetry.

Height, h cmh < 160160 ≤ h < 170170 ≤ h < 180h ≥ 180
Observed28708022

Heights of 200 adults

Degrees of freedom

For a uniform model whose end points are estimated from the data, subtract those too. Combine classes with E < 5 first.

Worked example

Test at the 5% level whether the heights fit N(170, 82).

  1. H0: N(170, 82) is a suitable model; H1: it is not.
  2. E = 21.13, 78.87, 78.87, 21.13
  3. X2 = 3.2833
  4. ν = 4 - 1 = 3 (no parameters estimated); critical value 7.815

Answer: 3.283 < 7.815: do not reject H0; N(170, 82) is a suitable model for the heights.

Key idea: Uniform: E proportional to class width. Normal: E = N × class probability with open end classes. Subtract one degree of freedom per estimated parameter.

Check yourself

  1. 120 arrivals are modelled as uniform over 0 to 60 minutes. Find the expected frequency for the class 20 to 40 minutes.
    Show answer

    40. 120 × 20/60 = 40.

  2. Arrivals in classes 0 to 10, 10 to 20, 20 to 40, 40 to 60 minutes are 26, 15, 37, 42. Calculate X2 for a uniform model on [0, 60]. Give your answer to 3 significant figures.
    Show answer

    3.38. E = 20, 20, 40, 40; X2 = 1.8 + 1.25 + 0.225 + 0.1 = 3.375.

  3. Heights of 200 adults are modelled by N(170, 82). Find the expected frequency for h < 160. Give your answer to 3 significant figures.
    Show answer

    21.1. 200 × Φ(-1.25) = 200 × 0.1056 = 21.130.

  4. Observed heights 28, 70, 80, 22 have expected frequencies 21.13, 78.87, 78.87, 21.13. Calculate X2. Give your answer to 3 significant figures.
    Show answer

    3.28. Σ(O - E)2/E = 3.2833.

  5. A normal distribution is fitted to data grouped into 7 classes (after combining), with μ and σ2 both estimated from the data. How many degrees of freedom are there?
    Show answer

    4. 7 - 1 - 2 = 4.

  6. When fitting a normal model, why are the end classes taken as h < 160 and h ≥ 180 rather than 150 ≤ h < 160 and 180 ≤ h < 190?
    • So that the class probabilities total 1 and the expected frequencies total N
    • To make the classes equal width
    • Because the normal distribution stops at 150 and 190
    • To avoid estimating σ
    Show answer

    So that the class probabilities total 1 and the expected frequencies total N. The normal model gives some probability to every value, so the end classes must be open.

  7. A continuous uniform distribution on [0, 60] is tested using 4 classes. How many degrees of freedom are there?
    • 2
    • 1
    • 3
    • 4
    Show answer

    3. The end points are given, so ν = 4 - 1 = 3.

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Last updated 8 November 2026. Spotted a mistake? Tell us.