EXM.1.SL.TZ0.8

pestleMathematicsAISLPaper 1EXM· sl-4-9-normal-distribution-and-calculationssource ↗

In an effort to study the level of intelligence of students entering college, a psychologist collected data from 4000 students who were given a standard test. The predictive norms for this particular test were computed from a very large population of scores having a normal distribution with mean 100 and standard deviation of 10. The psychologist wishes to determine whether the 4000 test scores he obtained also came from a normal distribution with mean 100 and standard deviation 10. He prepared the following table (expected frequencies are rounded to the nearest integer):

Copy and complete the table, showing how you arrived at your answers.

[5]

Test the hypothesis at the 5% level of significance.

[6]

Markscheme / solution

To calculate expected frequencies, we multiply 4000 by the probability of each cell:

$p\left( {80.5 \leqslant X \leqslant 90.5} \right) = p\left( {\frac{{80.5 - 100}}{{10}} \leqslant Z \leqslant \frac{{90.5 - 100}}{{10}}} \right)$ (M1)

$= p\left( { - 19.5 \leqslant Z \leqslant - 0.95} \right)$

$= 0.1711 - 0.0256$

$= 0.1455$

Therefore, the expected frequency $= 4000 \times 0.1455$ (M1)

$\approx 582$ (A1)

Similarly: $p\left( {90.5 \leqslant X \leqslant 100.5} \right) = 0.5199 - 0.1711$

$= 0.3488$

Frequency $= 4000 \times 0.3488$

$\approx 1396$ (A1)

And $p\left( {100.5 \leqslant X \leqslant 110.5} \right) = 0.8531 - 0.5199$

$= 0.3332$

Frequency $= 4000 \times 0.3332$

$\approx 1333$ (A1)

[5 marks]

To test the goodness of fit of the normal distribution, we use the ${\chi ^2}$ distribution. Since the last cell has an expected frequency less than 5, it is combined with the cell preceding it. There are therefore 7 – 1 = 6 degrees of freedom. (C1)

${\chi ^2} = \frac{{{{\left( {20 - 6} \right)}^2}}}{6} + \frac{{{{\left( {90 - 96} \right)}^2}}}{{96}} + \frac{{{{\left( {575 - 582} \right)}^2}}}{{582}} + \frac{{{{\left( {1282 - 1396} \right)}^2}}}{{1396}} + \frac{{{{\left( {1450 - 1333} \right)}^2}}}{{1333}} + \frac{{{{\left( {499 - 507} \right)}^2}}}{{507}} + {\frac{{\left( {84 - 80} \right)}}{{80}}^2}$ (M1)

= 53.03 (A1)

H₀: Distribution is Normal with $\mu = 100$ and $\sigma = 10$ .

H₁: Distribution is not Normal with $\mu = 100$ and $\sigma = 10$ . (M1)

$\chi _{\left( {0.95{\text{,}}\,\,6} \right)}^2 = 14.07$

Since ${\chi ^2} = 53.0 > \chi _{{\text{critical}}}^2 = 14.07$ , we reject H₀ (A1)

Or use of p-value

Therefore, we have enough evidence to suggest that the normal distribution with mean 100 and standard deviation 10 does not fit the data well. (R1)

Note: If a candidate has not combined the last 2 cells, award (C0)(M1)(A0)(M1)(A1)(R1) (or as appropriate).

[6 marks]

Examiners’ report

[N/A]