IB exam questions for topic 12

a) 1. Mark the value $175$ cm on the axis, to the left of the mean ($180$ cm). 2. Shade the area under the curve to the left of $175$.

b) (i) 3. $175$ is $5$ cm below the mean. By the symmetry of the curve, the value that leaves the same probability to its right is the reflection of $175$ in the mean, $5$ cm above it: $$h=180+5=185\text{cm}.$$

(ii) 4. One standard deviation above the mean is $180+5=185$ cm. We look for $P(X>185)$, which by the previous part is equal to $P(X<175)$. 5. Using the GDC (normal cumulative distribution, $\mu =180$, $\sigma =5$): $P(X>185)=0.15865\dots \approx 0.159$.

Answer: a) region shaded to the left of $175$; b) (i) $h=185$ cm; (ii) $0.159$

Let $T$ be the arrival time (minutes after opening): $T\sim N(52,5^2)$.

a) (i) 1. “At least $60$ minutes” is $P(T\ge 60)$. Using the GDC, normal cdf with mean $52$, standard deviation $5$, between $60$ and a very large value (for example $10^{99}$): $$P(T\ge 60)=0.05479\dots \approx \mathrm{0.0548.}$$

(ii) 2. Using the GDC, between $45$ and $55$: $$P(45<T<55)=0.6450\dots \approx \mathrm{0.645.}$$

b) 3. At Mulberry Park the distribution is the same, so the proportion of students who arrive at least $60$ minutes late is $0.05479\dots$. If $N$ is the number of students at the school: $$0.05479N=15\Rightarrow N=\frac{15}{0.05479\dots }=273.7\dots$$ 4. The number of students must be a whole number: approximately $274$.

Answer: a) (i) $0.0548$; (ii) $0.645$; b) $\approx 274$ students

Let $H\sim N(3.42,0.21^2)$ be the height, in metres.

a) 1. The mean is the centre of symmetry of the curve, so half of the area lies above it: $$P(H>3.42)=\mathrm{0.5.}$$

b) 2. Being within $2$ standard deviations of the mean means $\mu -2\sigma <H<\mu +2\sigma$, that is, $3.00<H<3.84$. 3. Using the GDC (normal cumulative distribution): $P(3.00<H<3.84)=0.95450\dots \approx 0.954$ (in agreement with the 95% rule).

c) 4. Using the GDC (normal cumulative distribution with lower bound $3.35$ and no upper bound): $P(H>3.35)=0.63055\dots \approx 0.631$.

d) 5. $P(H<x)=0.65$. Using the GDC, inverse normal with area to the left $0.65$, $\mu =3.42$ and $\sigma =0.21$: $$x=3.5009\dots \approx 3.50\text{m}.$$

Answer: a) $0.5$; b) $0.954$; c) $0.631$; d) $x\approx 3.50$ m

Let $T\sim N(34,3^2)$ be the daily temperature, in ${}^{\circ }\text{C}$.

a) 1. Using the GDC (normal cumulative distribution): $P(T<39)=0.95221\dots \approx 0.952$.

b) 2. The probability that a day exceeds $39{}^{\circ }\text{C}$ is the complement: $1-0.95221\dots =0.04779\dots$ 3. January has $31$ days, so the expected number of days is $$31\cdot 0.04779\dots =1.48\dots \approx 1.48\text{days}.$$

c) 4. $P(T>t)=0.7$ is equivalent to $P(T<t)=0.3$. 5. Using the GDC, inverse normal with area to the left $0.3$, $\mu =34$ and $\sigma =3$: $$t=32.43\dots \approx 32.4{}^{\circ }\text{C}.$$

Answer: a) $0.952$; b) $\approx 1.48$ days; c) $T\approx 32.4{}^{\circ }\text{C}$

Let $X$ be the time (in minutes) taken to complete the race: $X\sim N(28,5^2)$.

a) 1. The normal distribution is symmetric about its mean, which is $28$. Therefore half of the runners take more than $28$ minutes: $$P(X>28)=\mathrm{0.5.}$$

b) 2. Using the GDC, normal cdf with mean $28$, standard deviation $5$, lower limit very small (for example $-10^{99}$) and upper limit $26$: $$P(X<26)=0.3446\dots \approx \mathrm{0.345.}$$

c) 3. We know that $P(28<X<k)=0.20$. Since $P(X<28)=0.5$, it follows that $$P(X<k)=0.5+0.20=\mathrm{0.70.}$$ 4. Using the GDC, inverse normal with area to the left $0.70$, mean $28$ and standard deviation $5$: $$k=30.62\dots \approx 30.6\text{ minutes}.$$

Answer: a) $0.5$; b) $0.345$; c) $k=30.6$ minutes

Let $X$ be the mark: $X\sim N(300,100^2)$.

a) 1. 10% of the students obtain more than $k$: $P(X>k)=0.10$, that is, $P(X<k)=0.90$. 2. Using the GDC, inverse normal with area to the left $0.90$, mean $300$ and standard deviation $100$: $$k=428.16\dots \approx 428.$$

b) 3. Using the GDC, normal cdf with mean $300$, standard deviation $100$, between $450$ and a very large value: $$P(X\ge 450)=0.06681\dots \approx \mathrm{0.0668.}$$

c) 4. If Naomi studies at Marron University, her mark was at least $450$. We need a conditional probability: $$P(X\ge 500\mid X\ge 450)=\frac{P(X\ge 500)}{P(X\ge 450)},$$ because $X\ge 500$ implies $X\ge 450$. 5. Using the GDC, $P(X\ge 500)=0.02275\dots$ and $P(X\ge 450)=0.06681\dots$ (without rounding): $$\frac{0.02275\dots }{0.06681\dots }=0.3405\dots \approx \mathrm{0.341.}$$

Answer: a) $k=428$; b) $0.0668$; c) $0.341$

Let $X$ be the mass (g): $X\sim N(400,50^2)$.

a) 1. The mean is $400$ and the normal distribution is symmetric about it: $$P(X>400)=\mathrm{0.5.}$$

b) 2. 25% of the cobs exceed $a$: $P(X>a)=0.25$, that is, $P(X<a)=0.75$. 3. Using the GDC, inverse normal with area to the left $0.75$, mean $400$ and standard deviation $50$: $$a=433.72\dots \approx 434\text{ g}.$$

c) 4. The third quartile is the value that leaves $75\%$ of the data to its left: $Q_3=a=433.72\dots$ g. 5. By symmetry, the first quartile is $Q_1=400-(433.72\dots -400)=366.27\dots$ g (this also comes out of the inverse normal with area $0.25$). 6. Interquartile range: $$Q_3-Q_1=433.72\dots -366.27\dots =67.4\dots \approx 67.4\text{ g}.$$

Answer: a) $0.5$; b) $a\approx 434$ g; c) $67.4$ g

Let $T$ be the time taken (in minutes): $T\sim N(53,16.3^2)$.

a) 1. "At least 40 minutes" means $P(T\ge 40)$. Using the GDC, with the normal cumulative distribution function, mean $53$, standard deviation $16.3$, and bounds $40$ and a very large value: $$P(T\ge 40)=0.7874\dots \approx \mathrm{0.787.}$$

b) 2. 11% of the candidates took more than $k$ minutes: $P(T>k)=0.11$, so $P(T<k)=0.89$. 3. Using the GDC, the inverse normal with area to the left $0.89$, mean $53$ and standard deviation $16.3$: $$k=72.99\dots \approx 73.0\text{ minutes}.$$

c) 4. Using the GDC, $P(T<25)=0.04292\dots$. 5. Expected number out of $400$ candidates: $$400\cdot 0.04292\dots =17.2\dots \approx 17\text{ candidates}.$$

Answer: a) $0.787$; b) $k=73.0$ minutes; c) $\approx 17$ candidates

Let $W$ be the weight of a bag (g): $W\sim N(1000,4^2)$.

a) 1. Using the GDC, with the normal cumulative distribution function, mean $1000$, standard deviation $4$, between $990$ and $1004$: $$P(990<W<1004)=0.8351\dots \approx \mathrm{0.835.}$$

b) 2. 95% of the bags weigh less than $k$: $P(W<k)=0.95$. 3. Using the GDC, the inverse normal with area to the left $0.95$, mean $1000$ and standard deviation $4$: $$k=1006.58\dots \text{g}(\approx 1010\text{ g to 3 significant figures}).$$

c) 4. $P(1000-a<W<1000+a)=0.9$ is the symmetric interval around the mean that leaves a tail of $0.05$ on each side. Therefore $P(W<1000+a)=0.95$. 5. This is the same value as in part b): $1000+a=1006.58\dots$, so $$a=6.58\dots \approx 6.58\text{ g}.$$

Answer: a) $0.835$; b) $k=1006.6$ g ($\approx 1010$ g to 3 significant figures); c) $a=6.58$ g

a) 1. The normal curve is symmetric about its maximum, and the maximum is above $5800$. 2. In a normal distribution, the mean coincides with the central value of the curve: the mean useful life is $5800$ hours.

b) 3. We have $L\sim N(5800,850^2)$. 4. Using the GDC, normal cumulative distribution with lower bound $5000$, upper bound $6000$, $\mu =5800$ and $\sigma =850$: $$P(5000\le L\le 6000)=0.41970\dots \approx \mathrm{0.420.}$$

c) 5. We look for $k$ such that $P(L\ge k)=0.9$, that is, $P(L<k)=0.1$. 6. Using the GDC, inverse normal with area to the left $0.1$, $\mu =5800$ and $\sigma =850$: $$k=4710.68\dots$$ 7. To the nearest hundred, $k\approx 4700$ hours.

Answer: a) $5800$ hours; b) $0.420$; c) $k\approx 4700$ hours

a) 1. The centre of the bell indicates the mean, and its maximum height indicates the standard deviation: the smaller $\sigma$ is, the taller and narrower the curve (the maximum is approximately $\frac{0.4}{\sigma }$). 2. Mean $=-2$: the curve is centred at $x=-2$, that is, graph C. Check: $0.4/0.707\approx 0.57$, the height seen in C. 3. Mean $=0$ and $\sigma =0.447$: the curve is centred at $0$ and is tall, with maximum $0.4/0.447\approx 0.89$: graph D.

Let $X$ be the weight of a suitcase (kg): $X\sim N(20,3.5^2)$.

b) 4. Using the GDC, normal cdf with mean $20$, standard deviation $3.5$, upper limit $15$: $$P(X<15)=0.07656\dots \approx \mathrm{0.0766.}$$

c) 5. Excess baggage: $P(X>k)=0.196$, that is, $P(X<k)=1-0.196=0.804$. 6. Using the GDC, inverse normal with area to the left $0.804$, mean $20$ and standard deviation $3.5$: $$k=22.996\dots \approx 23.0\text{ kg}.$$

Answer: a) C and D (in that order); b) $0.0766$; c) $k=23$ kg

Let $X$ be the price (in euros) of one kilogram of tomatoes: $X\sim N(3.22,0.84^2)$.

a) (i) 1. The probability $P(X>3.22)$ corresponds to the region under the curve to the right of the mean, $3.22$ (the centre of the bell). We shade this right-hand half, from the vertical line drawn at $3.22$ to the end of the right-hand tail.

(ii) 2. Two standard deviations above the mean: $$\mu +2\sigma =3.22+2\cdot 0.84=3.22+1.68=4.90\text{ euros}.$$

b) 3. Using the GDC, normal cdf with mean $3.22$, standard deviation $0.84$, between $2.00$ and $3.00$: $$P(2<X<3)=0.3235\dots \approx \mathrm{0.323.}$$

c) 4. The sellers with a licence are the lowest $20\%$ of prices, so we look for $k$ such that $P(X<k)=0.20$. 5. Using the GDC, inverse normal with area to the left $0.20$, mean $3.22$ and standard deviation $0.84$: $$k=2.513\dots \approx 2.51\text{ euros per kg}.$$

Answer: a) (i) region to the right of $3.22$ shaded; (ii) $4.90$; b) $0.323$; c) $2.51$ euros per kg

a) 1. Using the GDC, we enter the grades as values and the frequencies as the frequency list, and request the one-variable statistics ($n=50$).

(i) 2. Mean: $\bar{x}=\frac{1\cdot 1+2\cdot 3+3\cdot 7+4\cdot 13+5\cdot 11+6\cdot 10+7\cdot 5}{50}=\frac{230}{50}=4.6$.

(ii) 3. Population standard deviation (the one the calculator gives as ${\sigma }_x$): $\sigma =1.456\dots \approx 1.46$.

b) 4. Cumulative frequencies: $1,4,11,24,35,45,50$. With $50$ data values, the median lies between the $25$th and the $26$th values. Both are in grade $5$ (the cumulative frequency goes from $24$ to $35$), so the median is $5$.

c) 5. First quartile: position $\frac{50}{4}=12.5$; the cumulative frequency first exceeds $12.5$ at grade $4$ (cumulative frequency $24$), so $Q_1=4$. 6. Third quartile: position $\frac{3\cdot 50}{4}=37.5$; the cumulative frequency first exceeds $37.5$ at grade $6$ (cumulative frequency $45$), so $Q_3=6$. 7. Interquartile range: $Q_3-Q_1=6-4=2$.

d) 8. Grades of $5$ or more: $11+10+5=26$ students. $$P(\text{5 or more})=\frac{26}{50}=\mathrm{0.52.}$$

e) 9. A second student is chosen from the same $50$, without replacing the first (they are two different students). Given that the first student has a grade of $5$ or more, there are $26$ possible cases for that student, and in $10$ of these cases the first student has a $6$. 10. Probability that the first student has a $6$, given that they have $5$ or more: $\frac{10}{26}$. 11. Then $49$ students remain, of whom $9$ have a $6$: $$P=\frac{10}{26}\cdot \frac{9}{49}=\frac{90}{1274}=0.0706\dots \approx \mathrm{0.0706.}$$

f) Let $M$ be the preparation time (minutes): $M\sim N(105,20^2)$.

(i) 12. Using the GDC, normal cdf with mean $105$, standard deviation $20$, between $90$ and a very large value: $$P(M\ge 90)=0.7733\dots \approx \mathrm{0.773.}$$

(ii) 13. Expected number among $50$ students: $$50\cdot 0.7733\dots =38.66\dots \approx \mathrm{38.7.}$$

Answer: a) (i) $4.6$; (ii) $1.46$; b) $5$; c) $2$; d) $0.52$; e) $0.0706$; f) (i) $0.773$; (ii) $38.7$ students

Let $W$ be the weight (kg): $W\sim N(65,5^2)$.

a) (i) 1. Using the GDC, normal cdf with mean $65$, standard deviation $5$, upper limit $61$: $$P(W<61)=0.2118\dots \approx \mathrm{0.212.}$$

(ii) 2. Expected number among $40$ players: $$40\cdot 0.2118\dots =8.47\dots \approx 8.48\text{ players}.$$

b) (i) 3. “Within $1.5$ standard deviations of the mean” means between $\mu -1.5\sigma =65-7.5=57.5$ and $\mu +1.5\sigma =65+7.5=72.5$. Draw a bell curve, mark these two values, symmetric about the centre, and shade the region between them.

(ii) 4. Using the GDC, normal cdf between $57.5$ and $72.5$ (mean $65$, standard deviation $5$): $$q=P(57.5<W<72.5)=0.8664\dots \approx \mathrm{0.866.}$$

c) 5. $P(W>k)=0.225$ is equivalent to $P(W<k)=1-0.225=0.775$. 6. Using the GDC, inverse normal with area to the left $0.775$, mean $65$ and standard deviation $5$: $$k=68.77\dots \approx 68.8\text{ kg}.$$

d) (i) 7. Null hypothesis: $H_0$: performance is independent of weight (and the alternative $H_1$: they are not independent).

(ii) 8. Using the GDC, ${\chi }^2$ test for independence on the observed matrix $$\left(\begin{array}{cc}6& 10\\ 7& 15\\ 12& 10\end{array}\right)$$ (a $3\times 2$ table, so the degrees of freedom are $(3-1)(2-1)=2$). The GDC gives ${\chi }_{\text{calc}}^2=2.49$ and the $p$-value: $$p=0.2874\dots \approx \mathrm{0.287.}$$ All the expected frequencies are greater than $5$ (the smallest is $16\cdot 25/60=6.67$), so the test is valid.

e) 9. We compare the $p$-value with the significance level: $0.287>0.05$, so we do not reject $H_0$. 10. Conclusion in context: there is not enough evidence, at the $5\%$ level, to claim that players' performance depends on their weight; the data are consistent with the two variables being independent.

Answer: a) (i) $0.212$; (ii) $8.48$ players; b) (i) region between $57.5$ and $72.5$ shaded; (ii) $q=0.866$; c) $k=68.8$ kg; d) (i) $H_0$: performance is independent of weight; (ii) $p=0.287$; e) do not reject $H_0$ because $p>0.05$ (there is no evidence of dependence between performance and weight)

a) 1. Draw a symmetric bell curve centred on the mean $67$, with marks at $60$, $67$ and $70$. 2. Shade the region under the curve between $x=60$ and $x=70$.

b) Let $V$ be the speed, with $V\sim N(67,3.4^2)$. We use the GDC (normal cumulative distribution).

(i) 3. $P(V>60)=0.98024\dots \approx 0.980$.

(ii) 4. $P(V<70)=0.81120\dots \approx 0.811$.

(iii) 5. $P(60<V<70)=0.79145\dots \approx 0.791$.

c) 6. 19% exceed the limit: $P(V>s)=0.19$, that is, $P(V<s)=0.81$. 7. Using the GDC, inverse normal with area to the left $0.81$, $\mu =67$ and $\sigma =3.4$: $s=69.98\dots$ 8. Rounding to the nearest integer, $s=70{\text{km h}}^{-1}$.

d) (i) 9. Each fine is 65 USD, so the number of fines is $$\frac{14820}{65}=228.$$

(ii) 10. These $228$ fines correspond to $19\%$ of the vehicles. If $n$ is the total number of vehicles: $$0.19n=228⟹n=\frac{228}{0.19}=1200.$$

Answer: a) diagram with the region between $60$ and $70$ shaded; b) (i) $0.980$; (ii) $0.811$; (iii) $0.791$; c) $s=70$; d) (i) $228$ fines; (ii) $1200$ vehicles

a) 1. Draw a symmetric bell curve with its maximum at the mean, $502$ g. 2. As $\sigma =2$, mark $500$ and $504$ (one standard deviation from the mean) and, further out, $498$ and $506$.

b) Let $X\sim N(502,2^2)$ be the weight of a box.

(i) 3. A box is sold if $497.5\le X\le 505$. Using the GDC (normal cumulative distribution): $$P(497.5\le X\le 505)=0.92097\dots \approx \mathrm{0.921.}$$

(ii) 4. Each day $1500$ boxes are produced, so we expect to sell $1500\cdot 0.92097\dots$ boxes, at $2 each: $$\text{Expected income}=1500\cdot 0.92097\dots \cdot 2\approx \$2760.$$

c) 5. The boxes that are not sold are recycled, with probability $1-0.92097\dots =0.07903\dots$ 6. Expected number of boxes recycled per day: $1500\cdot 0.07903\dots \approx 118.5$. 7. Cost of recycling: $$118.5\cdot 0.16\approx \$19.0.$$

d) 8. Let $Y\sim N(350,1.8^2)$. 97% are sold above $w$: $P(Y>w)=0.97$, so $P(Y<w)=0.03$. 9. Using the GDC, inverse normal with area to the left $0.03$, $\mu =350$ and $\sigma =1.8$: $$w=346.6\dots \approx 347\text{g}.$$

Answer: a) diagram; b) (i) $0.921$; (ii) $\approx \$2760$; c) $\approx \$19.0$; d) $w\approx 347$ g

a) 1. Using the GDC, enter the diameters as data and the frequencies as the frequency list, and obtain the one-variable statistics ($n=100$).

(i) 2. Mean: $\bar{x}=\frac{5\cdot 15+6\cdot 27+7\cdot 33+8\cdot 17+9\cdot 8}{100}=\frac{676}{100}=6.76$ cm.

(ii) 3. Standard deviation (population, ${\sigma }_x$): $1.14\dots \approx 1.14$ cm.

b) 4. Let $D\sim N(7,1.2^2)$. An apple is small if $D<6.5$. 5. Using the GDC: $P(D<6.5)=0.33846\dots$, that is, $33.8\%$ are small.

c) 6. 5% are large: $P(D\ge a)=0.05$, that is, $P(D<a)=0.95$. 7. Using the GDC, inverse normal with area to the left $0.95$, $\mu =7$ and $\sigma =1.2$: $$a=8.97\dots \approx 8.97\text{cm}.$$

d) 8. The medium apples are those that are neither small nor large: $$P(6.5\le D<a)=1-0.33846\dots -0.05=0.61154\dots$$ 9. The percentage of medium apples is $61.2\%$ (using the unrounded values: $61.15\dots \%$, which to 3 significant figures is $61.2\%$).

e) 10. 5% of $100000$ apples are large: $$0.05\cdot 100000=5000.$$

Answer: a) (i) $6.76$ cm; (ii) $1.14$ cm; b) $33.8\%$; c) $a\approx 8.97$ cm; d) $\approx 61.2\%$; e) $5000$ large apples

Let $W\sim N(110,8^2)$ be the weight, in kg.

a) 1. Draw a symmetric bell curve centred on $110$. 2. One standard deviation below the mean is $110-8=102$ and above it $110+8=118$; label $102$, $\mu =110$ and $118$.

b) (i) 3. The difference from the mean is $110-94=16$ kg, and since $\sigma =8$ kg: $$\frac{16}{8}=2\text{standard deviations below the mean}.$$

(ii) 4. Using the GDC (normal cumulative distribution): $P(W>94)=0.97725\dots \approx 0.977$.

c) (i) 5. Using the GDC: $P(88<W<116)=0.77039\dots \approx 0.770$.

(ii) 6. Expected number of sheep between $88$ and $116$ kg: $160\cdot 0.77039\dots =123.26\dots \approx 123$ sheep.

d) 7. $P(W<w)=0.75$. Using the GDC, inverse normal with area to the left $0.75$, $\mu =110$, $\sigma =8$: $$w=115.40\dots \approx 115\text{kg}.$$

e) 8. The two tails each have $30\%$, so the central region has $40\%$. 9. For $a$: $P(W<a)=0.3$. Inverse normal with area $0.3$: $a=105.8\dots \approx 106$. 10. For $b$: $P(W<b)=0.7$ (the area to its right is $0.3$). Inverse normal with area $0.7$: $b=114.2\dots \approx 114$.

Answer: a) diagram with $102$, $110$ and $118$; b) (i) $2$ standard deviations; (ii) $0.977$; c) (i) $0.770$; (ii) $\approx 123$ sheep; d) $w\approx 115$ kg; e) $a\approx 106$, $b\approx 114$

Let $Q\sim N(60,12^2)$ be the Chemistry score.

a) 1. Draw a symmetric bell curve centred on $60$ (the maximum is above $60$), indicating the standard deviation $12$ as the horizontal distance from the axis of symmetry to the curve.

b) 2. As $60$ is the mean and the curve is symmetric, $P(Q\le 60)=0.5$.

c) 3. Using the GDC (normal cumulative distribution with lower bound $80$): $P(Q>80)=0.04779\dots \approx 0.0478$.

d) 4. Let $F\sim N(63,10^2)$ be the Physics score. Using the GDC: $P(F<80)=0.95543\dots \approx 0.955$.

e) 5. In Chemistry, Hee Jin scores higher than $1-0.0478=0.952$ of the candidates ($95.2\%$). In Physics, Hee Jin scores higher than $0.955$ ($95.5\%$). 6. As $0.955>0.952$, the Physics score is better compared with the other candidates, because it is higher than a greater proportion of them.

f) 7. Being in the top $10\%$ means $P(F\ge x)=0.10$, that is, $P(F<x)=0.9$. 8. Using the GDC, inverse normal with area to the left $0.9$, $\mu =63$ and $\sigma =10$: $x=75.8\dots$ 9. As the minimum score must be an integer and must lie within the top $10\%$, the value rounded to the nearest integer is $76$.

Answer: a) diagram; b) $0.5$; c) $0.0478$; d) $0.955$; e) better in Physics (higher than more candidates); f) $76$

a) 1. Substitute $t=6$ into the model: $$N=200\cdot 1.9^{0.85\cdot 6}=200\cdot 1.9^{5.1}=5280.48\dots$$ 2. There will be about 5,280 infected computers.

b) 3. We set up $200\cdot 1.9^{0.85t}>1000000$, and solve: $$1.9^{0.85t}>5000.$$ 4. Taking logarithms (or solving the equation $200\cdot 1.9^{0.85t}=1000000$ with the GDC): $$0.85t>\frac{\ln 5000}{\ln 1.9}⟹t>\frac{\ln 5000}{0.85\ln 1.9}=15.61\dots$$ 5. Rounding to the nearest whole number of hours, $t=16$ hours.

Answer: a) $\approx 5280$ computers; b) $t=16$ hours