IB Maths AA SL 4 Statistics and Probability Question Bank

2 (days)

using expected value formula
(mean =) 2.44 (2.436 (exact))

convenience (sampling)

recognizing that the median is at half the total frequency
(M1)
eg $\frac{2000}{2}$
m=2500 (dollars)

500 families have a monthly income less than 2000

A1
N1

correct cumulative frequency, 1850
subtracting their cumulative frequency from 2000
eg 2000-1850
150 families have a monthly income of more than 4000 dollars

Note: If working shown, award M1A1A1 for 128+22=150, using the table.

correct calculation
eg 2000-(436+64+765+28+122), 1850-500-765
p=585

evidence of using correct mid-interval values (1500,3000,4500)
attempt to substitute into $\frac{\sum f x}{\sum f}$
eg $\frac{1500 \times 64+3000 \times p+4500 \times 122}{64+585+122}$
3112.84

3110 [3110, 3120] (dollars)

attempt to substitute F=19.8 into the regression line for A on F
A=2.89(19.8)+99.3
$=156.522(\mathrm{~cm})$
arm span $=157(\mathrm{~cm})$

Note: Award M0A0 for choosing the wrong regression line to get
$A=156.417 \ldots$ so A=156.

recognition that the lines intersect at the mean point (may be seen on a sketch)
$2.89 F+99.3=\frac{F+32.6}{0.335}$ OR $0.335 A-32.6=\frac{A-99.3}{2.89}$
159.686… or $20.8948 \ldots$
the mean arm span $=160(\mathrm{~cm})$, the mean foot length $=20.9(\mathrm{~cm})$

METHOD 1
recognition of symmetry of interval around mean (may be seen on a sketch)
P(H<153)=0.06 OR P(H<173)=0.94 OR equivalent
$\frac{153-163}{\sigma}=-1.55477 \ldots$ OR $\frac{173-163}{\sigma}=1.55477 \ldots$$\sigma=6.43181 \ldots$$\sigma=6.43(\mathrm{~cm})$

METHOD 2
attempt to find $\sigma$ by equating an appropriate correct normal CDF function to 0.88
(or e.g. 0.06 or 0.94 )
$\sigma=6.43181 \ldots$$\sigma=6.43(\mathrm{~cm})$

Note: Accept use of calculator notation e.g.
normcdf $(153,173,163, \sigma)=0.88$

0.010724
0.0107

correct z-value
0.263714...
evidence of appropriate approach
eg $\quad \frac{0.65-0.592}{\sigma}, 0.264=\frac{x-\mu}{\sigma}$
correct substitution
eg $\quad 0.263714=\frac{0.65-0.592}{\sigma}, \sigma=\frac{0.65-0.592}{0.264}$
0.219934
$\sigma=0.220$

correct work for P (group X and t>0.65 ) or P (group Y and t>0.65 ) (may be seen anywhere)
eg P( group X$) \times \mathrm{P}(t>0.65 \mid \mathrm{X}), \mathrm{P}(X \cap t>0.65)=0.0107 \times 0.38(=0.004075)$, $\mathrm{P}(Y \cap t>0.65)=0.396 \times 0.62$
recognizing conditional probability (seen anywhere)
eg $\mathrm{P}(\mathrm{X} \mid t>0.65), \mathrm{P}(A \mid B)=\frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$
valid approach to find P(t>0.65)

correct work for P(t>0.65)
eg $\quad 0.0107 \times 0.38+0.396 \times 0.62,0.249595$
correct substitution into conditional probability formula
eg $\frac{0.0107 \times 0.38}{0.0107 \times 0.38+0.396 \times 0.62}, \frac{0.004075}{0.249595}$
0.016327
$\mathrm{P}(\mathrm{X} \mid t>0.65)=0.0163270$

recognizing binomial probability
eg $\quad X \sim \mathrm{~B}(n, p),\binom{n}{r} p^{r} q^{n-r},(0.016327)^{2}(0.983672)^{8},\binom{10}{2}$
valid approach
eg $\mathrm{P}(X \geq 2)=1-\mathrm{P}(X \leq 1), 1-\mathrm{P}(X<a)$, summing terms from 2 to 10 (accept binomcdf( 10,0.0163,2,10 ) )
0.010994
$\mathrm{P}(X \geq 2)=0.0110$