Limits, Probability, and Statistics: practice solutions, Grade 12 – download the PDF

1 Reading a limit from a table ★★★

For \( x\neq4 \), \( f(x)=\dfrac{(x-4)(x+4)}{x-4}=x+4 \). So \( f(3.9)=7.9 \), \( f(3.99)=7.99 \), \( f(4.01)=8.01 \), \( f(4.1)=8.1 \).

The values approach 8 from both sides, so \( \lim_{x\to4}f(x)=8 \), even though \( f(4) \) is undefined.

2 One-sided limits of a piecewise function ★★★

From the left use \( x+2 \): \( 1+2=3 \). From the right use \( 6-x \): \( 6-1=5 \).

Since \( 3\neq5 \), the two-sided limit does not exist.

3 Two limits at infinity ★★★

1. The degrees are equal, so the limit is the ratio of leading coefficients: \( \dfrac51=5 \).
2. The numerator is constant while the denominator grows without bound, so the limit is 0.

4 Podium or committee? ★★★

1. Order matters: \( P(6,2)=6\cdot5=30 \).
2. Order does not matter: \( C(6,2)=\dfrac{6\cdot5}{2}=15 \).

Check: \( P(6,2)=2!\cdot C(6,2)=30 \).

5 Marbles without replacement ★★★

The first is red with probability \( \dfrac58 \). Given that, 4 red remain among 7, so \( P(\text{2nd red}\mid\text{1st red})=\dfrac47 \).

\( P=\dfrac58\cdot\dfrac47=\dfrac{20}{56}=\dfrac{5}{14}\approx0.357 \).

6 Coin flips ★★★

This is binomial with \( n=4 \), \( p=0.5 \): \( P(X=3)=C(4,3)(0.5)^3(0.5)^1=4\cdot\dfrac1{16}=0.25 \).

7 Computing z-scores ★★★

1. \( z=\dfrac{85-70}{10}=1.5 \).
2. \( x=\mu+z\sigma=70+(-2)(10)=50 \).

8 Limit by factoring ★★★

Substitution gives \( \tfrac00 \). Factor: \( x^2+x-12=(x+4)(x-3) \).

For \( x\neq3 \) the fraction equals \( x+4 \), so the limit is \( 3+4=7 \).

9 Three limits at infinity ★★★

1. Equal degrees: \( \dfrac42=2 \).
2. Degree 2 below degree 3: the limit is 0.
3. Degree 3 above degree 2: dividing by \( x^2 \) gives \( \dfrac{x}{1+4/x^2} \), which grows without bound, so there is no finite limit.

10 Arrangements of BANANA ★★★

There are 6 letters with A repeated 3 times and N repeated 2 times.

\( \dfrac{6!}{3!\,2!}=\dfrac{720}{12}=60 \) arrangements.

11 Choosing a committee ★★★

Choose the girls: \( C(6,3)=20 \). Choose the boys: \( C(5,2)=10 \).

By the counting principle, \( 20\cdot10=200 \) committees.

12 Laptops and tablets ★★★

\( P(\text{tablet}\mid\text{laptop})=\dfrac{P(\text{both})}{P(\text{laptop})}=\dfrac{0.45}{0.60}=0.75 \).

The probability is 75%.

13 Free throws, exactly four ★★★

\( n=5 \), \( p=0.7 \), \( k=4 \): \( P=C(5,4)(0.7)^4(0.3)^1=5\cdot0.2401\cdot0.3=0.36015 \).

The probability is about 0.360, or 36%.

14 Reading scores ★★★

1. 85 and 115 are \( 1\sigma \) from the mean, so about 68%.
2. 130 is \( 2\sigma \) above the mean. About 95% lie within \( 2\sigma \), so \( \dfrac{100-95}{2}=2.5\% \) lie above 130.

15 Standard deviation of a short list ★★★

Mean: \( \dfrac{12+15+15+18+20}{5}=16 \). Deviations squared: \( 16,1,1,4,16 \), sum \( 38 \).

Population: \( \sigma=\sqrt{38/5}=\sqrt{7.6}\approx2.76 \). Sample: \( s=\sqrt{38/4}=\sqrt{9.5}\approx3.08 \) minutes.

16 Limit with a conjugate ★★★

Substitution gives \( \tfrac00 \). Multiply top and bottom by the conjugate \( \sqrt{x+9}+3 \):

\( \dfrac{(x+9)-9}{x(\sqrt{x+9}+3)}=\dfrac{x}{x(\sqrt{x+9}+3)}=\dfrac1{\sqrt{x+9}+3} \) for \( x\neq0 \).

As \( x\to0 \) this tends to \( \dfrac1{3+3}=\dfrac16 \).

17 Asymptotes of a rational function ★★★

Factor: \( f(x)=\dfrac{3(x+2)}{(x-2)(x+1)} \). No factor cancels.

Vertical asymptotes: the denominator is 0 at \( x=2 \) and \( x=-1 \), where the numerator is not 0.

Horizontal asymptote: the numerator degree 1 is below the denominator degree 2, so \( \lim_{x\to\pm\infty}f(x)=0 \) and the asymptote is \( y=0 \).

18 PIN codes ★★★

First digit: 9 choices (1 to 9). Second digit: any unused digit, including 0, so 9 choices. Third: 8. Fourth: 7.

\( 9\cdot9\cdot8\cdot7=4536 \) PINs.

19 A medical screening test ★★★

\( P(\text{sick and +})=0.02\cdot0.95=0.019 \). \( P(\text{healthy and +})=0.98\cdot0.04=0.0392 \).

\( P(+)=0.019+0.0392=0.0582 \).

\( P(\text{sick}\mid +)=\dfrac{0.019}{0.0582}\approx0.326 \). Only about 32.6%, because the condition is rare.

20 Guessing on a quiz ★★★

\( n=10 \), \( p=0.25 \).

1. \( \mu=np=2.5 \).
2. \( \sigma=\sqrt{10\cdot0.25\cdot0.75}=\sqrt{1.875}\approx1.37 \).
3. \( P(X\ge1)=1-P(X=0)=1-0.75^{10}\approx1-0.0563=0.9437 \).

21 Bolts and tolerances ★★★

1. \( z=\dfrac{49.2-50}{0.4}=-2 \) and \( z=\dfrac{50.8-50}{0.4}=2 \): about 95%.
2. \( z=\dfrac{50.5-50}{0.4}=1.25 \).
3. \( z=\dfrac{49.4-50}{0.4}=-1.5 \). It lies within \( 2\sigma \) of the mean, so it is not unusual.

22 Price and sales ★★★

\( \bar x=4 \), \( \bar y=66 \). \( \sum(x-\bar x)(y-\bar y)=(-2)(24)+(-1)(9)+0\cdot4+(1)(-11)+(2)(-26)=-120 \). \( \sum(x-\bar x)^2=10 \). \( \sum(y-\bar y)^2=1470 \).

1. \( m=\dfrac{-120}{10}=-12 \), \( b=66-(-12)(4)=114 \), so \( \hat y=-12x+114 \).
2. \( r=\dfrac{-120}{\sqrt{10\cdot1470}}\approx-0.99 \): a very strong negative linear relationship.
3. Each extra dollar lowers predicted sales by about 12 units. At \( x=3.5 \), \( \hat y=-12(3.5)+114=72 \) units.