(a) Unequal intervals → Lagrange's formula gives $\approx108.75$ (thousand tonnes). (b) Poisson $\lambda=2$: $P(0)=e^{-2}=0.1353$; $P(X\ge3)=1-[P(0)+P(1)+P(2)]=0.3235$.

Part (a) — Lagrange's interpolation (unequal intervals)

Data: $(1974,25),(1978,60),(1982,80),(1990,170)$; estimate at $x=1986$. Since the arguments are not equally spaced, use Lagrange's formula.

Compute each Lagrange coefficient at $x=1986$:

$L_0=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}=\frac{(8)(4)(-4)}{(-4)(-8)(-16)}=\frac{-128}{-512}=0.25,$

$L_1=\frac{(12)(4)(-4)}{(4)(-4)(-12)}=\frac{-192}{192}=-1,$

$L_2=\frac{(12)(8)(-4)}{(8)(4)(-8)}=\frac{-384}{-256}=1.5,$

$L_3=\frac{(12)(8)(4)}{(16)(12)(8)}=\frac{384}{1536}=0.25.$

Interpolated value:

$y(1986)=25(0.25)+60(-1)+80(1.5)+170(0.25)$

$=6.25-60+120+42.5=108.75.$

Part (b) — Poisson distribution ($\lambda=2$)

$P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}$ with $\lambda=2,\ e^{-2}=0.1353$.

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