The number, X, of books received at a charity shop has a constant mean of 5.1 per day.
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] 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) Allow incorrect . Allow one end error. Must see an expression (accept correct sigma notation).
=0.945(3 sf) SC No working, 0.945 B1(could be implied) SC B1.
Question 2
[Maximum number: 3]
Cars arrive at a fuel station at random and at a constant average rate of 13.5 per hour.
Find the probability that the total number of cars and trucks arriving at the fuel station during a 10 -minute period is more than 3 and less than 7 .
=13.5/6+3.6 x 2/3 OE or 4.65 Attempt to find e^-4.65(4.65^44!+4.65^55!+4.65^66!) Allow any Allow one end error Poisson terms not be seen
0.494 (3 sf) If M0 allow SC B1 for 0.494
Question 3
[Maximum number: 5]
The random variable T denotes the time, in seconds, for 100 m races run by Tania. T is normally distributed with mean μ and variance σ2. A random sample of 40 races run by Tania gave the following results.
n=40Σt=560Σt2=7850
Using your answers to part (a), find the probability that, in a randomly chosen 100 m race, Suki's time will be at least 0.1 s more than Tania's time.
E(S-T)=14.2-14=0.2 B1 FT FT their 14.
Var(S−T)=0.3+0.256=0.55641 B1 FT Accept 390217. FT their 0.256, including FT biased. Var(S−T)=0.55.
0.556410.1−0.2=−0.134 Standardising with their values; biased gives -0.135. FT their mean and variance.
P(S−T>0.1)=1−Φ(−0.134)=Φ(0.134) Finding correct area consistent with their values.
0.553(3sf) Use of biased gives 0.554 (3 sf) and can score A1. Similar schemes for P(T-S)<-0.1, S-T-0.1>0, and T-S+0.1<0.