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Date November 2010 Marks available 15 Reference code 10N.3sp.hl.TZ0.2
Level HL only Paper Paper 3 Statistics and probability Time zone TZ0
Command term Find Question number 2 Adapted from N/A

Question

The length of time, T, in months, that a football manager stays in his job before he is removed can be approximately modelled by a normal distribution with population mean \(\mu \) and population variance \({\sigma ^2}\). An independent sample of five values of T is given below.

6.5, 12.4, 18.2, 3.7, 5.4

(a)     Given that \({\sigma ^2} = 9\),

(i)     use the above sample to find the 95 % confidence interval for \(\mu \), giving the bounds of the interval to two decimal places;

(ii)     find the smallest number of values of T that would be required in a sample for the total width of the 90 % confidence interval for \(\mu \) to be less than 2 months.

(b)     If the value of \({\sigma ^2}\) is unknown, use the above sample to find the 95 % confidence interval for \(\mu \), giving the bounds of the interval to two decimal places.

Markscheme

(a)     (i)     as \({\sigma ^2}\) is known \({\bar x}\) is \({\text{N}}\left( {\mu ,\frac{{{\sigma ^2}}}{n}} \right)\)     (M1)

CI is \(\bar x - {z^ * }\frac{\sigma }{{\sqrt n }} < \mu  < \bar x + {z^ * }\frac{\sigma }{{\sqrt n }}\)     (M1)

\(\bar x = 9.24,{\text{ }}{z^ * } = 1.960\) for 95 % CI     (A1)

CI is \(6.61 < \mu < 11.87\) by GDC     A1A1

 

(ii)     CI is \(\bar x - {z^ * }\frac{\sigma }{{\sqrt n }} < \mu < \bar x + {z^ * }\frac{\sigma }{{\sqrt n }}\)

require \(2 \times 1.645\frac{3}{{\sqrt n }} < 2\)     R1A1

\(4.935 < \sqrt n \)     (A1)

\(24.35 < n\)     A1

so smallest value for n = 25     A1

Note: Accept use of table.

 

[10 marks]

 

(b)     as \({\sigma ^2}\) is not known \({\bar x}\) has the t distribution with v = 4     (M1)(A1)

CI is \(\bar x - {t^ * }\frac{{{s_{n - 1}}}}{{\sqrt n }} < \mu < \bar x + {t^ * }\frac{{{s_{n - 1}}}}{{\sqrt n }}\)

\(\bar x = 9.24,{\text{ }}{s_{n - 1}} = 5.984,{\text{ }}{t^ * } = 2.776\) for 95 % CI     (A1)

CI is \(1.81 < \mu < 16.67\) by GDC     A1A1

[5 marks]

Total [15 marks]

Examiners report

The 2 confidence intervals were generally done well by using a calculator. Some marks were dropped by not giving the answers to 2 decimal places as required. Weak candidates did not realise that (b) was a t interval. Part (a) (ii) was not as well answered and often it was the first step that was the problem.

Syllabus sections

Topic 7 - Option: Statistics and probability » 7.5 » Confidence intervals for the mean of a normal population.

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