Date | May 2017 | Marks available | 2 | Reference code | 17M.2.sl.TZ1.9 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
A random variable XX is normally distributed with mean, μμ. In the following diagram, the shaded region between 9 and μμ represents 30% of the distribution.
The standard deviation of XX is 2.1.
The random variable YY is normally distributed with mean λλ and standard deviation 3.5. The events X>9X>9 and Y>9Y>9 are independent, and P((X>9)∩(Y>9))=0.4P((X>9)∩(Y>9))=0.4.
Find P(X<9)P(X<9).
Find the value of μμ.
Find λλ.
Given that Y>9Y>9, find P(Y<13)P(Y<13).
Markscheme
valid approach (M1)
egP(X<μ)=0.5, 0.5−0.3P(X<μ)=0.5, 0.5−0.3
P(X<9)=0.2P(X<9)=0.2 (exact) A1 N2
[2 marks]
z=−0.841621z=−0.841621 (may be seen in equation) (A1)
valid attempt to set up an equation with their zz (M1)
eg−0.842=μ−Xσ, −0.842=X−μσ, z=9−μ2.1−0.842=μ−Xσ, −0.842=X−μσ, z=9−μ2.1
10.7674
μ=10.8μ=10.8 A1 N3
[3 marks]
P(X>9)=0.8P(X>9)=0.8 (seen anywhere) (A1)
valid approach (M1)
egP(A)×P(B)P(A)×P(B)
correct equation (A1)
eg0.8×P(Y>9)=0.40.8×P(Y>9)=0.4
P(Y>9)=0.5P(Y>9)=0.5 A1
λ=9λ=9 A1 N3
[5 marks]
finding P(9<Y<13)=0.373450P(9<Y<13)=0.373450 (seen anywhere) (A2)
recognizing conditional probability (M1)
egP(A|B), P(Y<13|Y>9)P(A|B), P(Y<13|Y>9)
correct working (A1)
eg0.3730.50.3730.5
0.746901
0.747 A1 N3
[5 marks]