Q.(a) The mean score of 1000 students for an examination is 34 and the standard deviation is 16. Determine the limit of the marks of the central 70% of the candidates by assuming the distribution is normal. $P[0 < Z < 1.04] = 0.35$ OR

(a) uses the standard normal table value $P[0<Z<1.04]=0.35$ to locate the symmetric limits containing the central $70\%$ of scores; (b) splits $\int_0^\pi|\sin x-\cos x|\,dx$ at the crossing point $x=\pi/4$ and integrates each piece.

(a) Central $70\%$ limits, $\mu=34,\ \sigma=16$

1. For a normal distribution, the central $70\%$ of observations lie symmetrically about the mean, i.e. $35\%$ on each side of $\mu$ (since $70\%/2=35\%=0.35$).
2. We are given $P[0<Z<1.04]=0.35$, so the required $z$-limits are $Z=\pm1.04$, i.e. $P[-1.04<Z<1.04]=2(0.35)=0.70$.
3. Convert to raw scores using $X=\mu+Z\sigma$: lower limit $=34-1.04(16)=34-16.64=17.36$; upper limit $=34+1.04(16)=34+16.64=50.64$.
4. So the central $70\%$ of the 1000 candidates scored between $17.36$ and $50.64$ marks.

(b) Area between $y=\sin x$ and $y=\cos x$ on $[0,\pi]$

1. Find where the curves cross in $[0,\pi]$: $\sin x=\cos x\Rightarrow\tan x=1\Rightarrow x=\dfrac{\pi}{4}$ (the only solution in this interval).
2. On $\big[0,\tfrac{\pi}{4}\big]$, $\cos x\ge\sin x$ (e.g. at $x=0$: $\cos0=1>\sin0=0$), so the integrand is $\cos x-\sin x$.
3. On $\big[\tfrac{\pi}{4},\pi\big]$, $\sin x\ge\cos x$ (e.g. at $x=\pi/2$: $\sin=1>\cos=0$), so the integrand is $\sin x-\cos x$.
4. Area $=\displaystyle\int_0^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi}(\sin x-\cos x)\,dx$. …