Practise Probability & Statistics 2 through Poisson and normal models, continuous variables, sampling, confidence intervals and hypothesis tests with contextual conclusions.
The number, X, of books received at a charity shop has a constant mean of 5.1 per day.
State, in context, one condition for X to be modelled by a Poisson distribution.
Assume now that X can be modelled by a Poisson distribution.
Books received independently or singly or randomly.
B1
OE. Must be in context.
If more than one condition given, ignore extras.
Find the probability that exactly 10 books are received in a 3-day period.
e^-15.3 x 15.3^1010!
M1
Allow incorrect .
=0.0439(3 sf)
A1
SC No working shown but correct answer seen scores
B1.
Use a suitable approximating distribution to find the probability that more than 180 books are received in a 30-day period.
The number of DVDs received at the same shop is modelled by an independent Poisson distribution with mean 2.5 per day.
N(153, 153)
B1
Seen or implied.
180.5-153153 [=2.223]
M1
For standardising with their values (can be implied).
Allow with wrong or missing continuity correction.
1- ('2.223')
M1
For correct probability area consistent with their values.
=0.0131(3 sf)
A1
Find the probability that the total number of books and DVDs that are received at the shop in 1 day is more than 3 .
(=) 5.1+2.5 [=7.6]
B1
Give at early stage (seen or implied).
1-e^-7.6(1+7.6+7.6^22+7.6^33!)=1-e^-7.6(1+7.6+28.88+73.16) =1-(0.0005005+0.003803+0.01445+0.03661)1-e^-7.6(1+7.6+7.6^22+7.6^33!)=1-e^-7.6(1+7.6+28.88+73.16)=1-(0.0005005+0.003803+0.01445+0.03661)
M1
Allow incorrect . Allow one end error.
Must see an expression (accept correct sigma notation).
=0.945(3 sf)
A1
SC No working, 0.945 B1(could be implied) SC B1.
A random variable X has probability density function f given by
where a is a constant.
Show that a=2.
∫02(ax−x3)dx=1
M1
Attempted integration of f(x) and equated to 1.
[a2x2−4x4]02=1a−44=1
A1
Correct integration and substitute correct limits.
a=2
A1
AG Convincingly obtained and no errors seen.
Find the median of X.
∫0m(2x−x3)dx=
M1
Attempt integrate f(x) with limits 0 to m (or m to 2 ) and equate to 21.
m2−4m4=21
A1
For correct quartic in any form.
m4−4m2+2=0⇒m2=24±16−8[=2±2]
M1
For solving their three term quartic to find m2.
m=2−2 or 0.765(3 sf)
A1
Find the exact value of E(X).
∫02(2x2−x4)dx
M1
Attempt to integrate x f(x). Ignore limits.
[32x3−5x5]02
A1
Correct integration and correct limits.
[=342−542]=1582
A1
OE For single exact term.
Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries.
Give a reason for using a sample rather than the whole population in carrying out this test.
A random sample of 120 type B batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours 2.
Batteries unusable after testing or Population too big
or too costly or too time consuming to use the whole population oe
B1
Test, at the 2% significance level, whether type B batteries have a shorter mean life than type A batteries.
H_0: µ=150H_1: µ<150
B1
Or population mean =150; not just 'mean' =150
147-150225 / 120
M1
Allow with continuity correction
Need 120
-2.191
A1
Condone - 2.19
-2.191<-2.054 [or -2.055]
M1
OE. For valid comparison with 2.054 or 2.055 Or 0.0143 (or 0.0142) < 0.02
For two tail test allow comp -2.326 OE if H_1: µ != 150 (can score B0M1A1M1A0 max 3/5)
[Reject Ho] There is evidence that the (mean) life of type B is less than type
A (or less than 150)
A1 FT
In context, not definite with no contradictions
Accept critical value method 147.19 M1A1
147<147.19 M1 conclusion A1
Or 150 > 149.81
Calculate a 94% confidence interval for the population mean life of type B batteries.
147 +/- z x 15120
M1
Expression of correct form must be a z value
z=1.881 [or 1.882]
B1
144 to 150 (3 s.f.)
A1
Must be an interval
Incorrect z value can only score M1B0A0