Q.(a) In an investigation, a corpse was found by a detective at exactly 8 p.m. Being alert, the detective also measured the body temperature and found it to be 70°F. Two hours later, the detective measured the body temperature again and found it to be 60°F. If the room temperature is 50°F, and assuming that the body temperature of the person before death was 98.6°F, prove that the time of death is 5.26 p.m. (5 hrs 26 minutes) (app.). $\left[\dfrac{\log(2.43)}{\log(2)}\simeq1.28\right]$ OR

(a) applies Newton's law of cooling $T=T_{room}+(T_0-T_{room})e^{-kt}$ with two timed readings to find how long before 8 p.m. the body temperature was $98.6^\circ F$; (b) tabulates the probability mass function of the number of heads in 3 fair-coin tosses, computes its mean and variance directly, and checks them against the Binomial formulas $np,npq$.

(a) Time of death by Newton's law of cooling

1. Newton's law of cooling: $\dfrac{dT}{dt}=-k(T-T_{room})\Rightarrow T(t)=T_{room}+(T_0-T_{room})e^{-kt}$, where $t$ is measured from the (unknown) time of death and $T_0=98.6^\circ F$ is the body temperature at death, $T_{room}=50^\circ F$.
2. So $T(t)-50=48.6\,e^{-kt}$. Let $t_1$ = time elapsed from death to the 8 p.m. reading.
3. At 8 p.m., $T=70$: $70-50=20=48.6\,e^{-kt_1}\Rightarrow e^{-kt_1}=\dfrac{20}{48.6}$.
4. At 10 p.m. (i.e. $t_1+2$), $T=60$: $60-50=10=48.6\,e^{-k(t_1+2)}\Rightarrow e^{-k(t_1+2)}=\dfrac{10}{48.6}$.
5. Divide the two: $\dfrac{e^{-kt_1}}{e^{-k(t_1+2)}}=e^{2k}=\dfrac{20/48.6}{10/48.6}=2\Rightarrow e^{2k}=2\Rightarrow k=\dfrac12\ln2$.
6. From step 3: $-kt_1=\ln\left(\dfrac{20}{48.6}\right)=\ln\left(\dfrac{100}{243}\right)=-\ln\left(\dfrac{243}{100}\right)=-\ln(2.43)$, so $t_1=\dfrac{\ln(2.43)}{k}=\dfrac{\ln(2.43)}{\frac12\ln2}=\dfrac{2\ln(2.43)}{\ln2}=\dfrac{2\log(2.43)}{\log2}$ (the ratio is base-independent).
7. Using the given value $\dfrac{\log(2.43)}{\log2}\approx1.28$: $t_1\approx2(1.28)=2.56$ hr $=2$ hr $+0.56\times60$ min $\approx2$ hr $34$ min.
8. Time of death $=8{:}00$ p.m. $-\,2$ hr $34$ min $=5{:}26$ p.m., as required to prove.

(b) PMF, mean, variance for number of heads in 3 coin tosses

1. Sample space (8 equally likely outcomes): $HHH,HHT,HTH,THH,HTT,THT,TTH,TTT$. Let $X$ = number of heads, $X\in\{0,1,2,3\}$.
2. $P(X=0)=\dfrac18$ (TTT); $P(X=1)=\dfrac38$ (HTT, THT, TTH); $P(X=2)=\dfrac38$ (HHT, HTH, THH); $P(X=3)=\dfrac18$ (HHH).
3. Mean: $E(X)=\sum xP(x)=0\left(\dfrac18\right)+1\left(\dfrac38\right)+2\left(\dfrac38\right)+3\left(\dfrac18\right)=\dfrac{0+3+6+3}{8}=\dfrac{12}{8}=\dfrac32$. …