Q.The mean score of 1000 students for an examination is 34 and S.D. is 16.

Standardise the score limits to $z$-values, use the given normal-table areas to find the proportion (and hence the count) between 30 and 60, then invert to find the central-70% limits.

1. Given data. $N=1000$ students, mean $\mu=34$, standard deviation $\sigma=16$. $Z=\dfrac{X-\mu}{\sigma}$.

2. Part (i): candidates scoring between 30 and 60.

   For $X=30$: $z_1=\dfrac{30-34}{16}=\dfrac{-4}{16}=-0.25$

   For $X=60$: $z_2=\dfrac{60-34}{16}=\dfrac{26}{16}=1.625\approx1.63$

   $P(30<X<60)=P(-0.25<Z<1.63)=P(0<Z<0.25)+P(0<Z<1.63)$ (splitting at the mean, using symmetry of the normal curve)

   $=0.0987+0.4484=0.5471$

   Expected number of candidates $=N\times P=1000\times0.5471=547.1\approx547$

3. Part (ii): limits of the central 70% of candidates.

   The central 70% is the symmetric interval $\mu-z_0\sigma$ to $\mu+z_0\sigma$ such that $P(-z_0<Z<z_0)=0.70$.

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