Answer
The middle 99% of bags have between 958 and 1566 chocolate chips.
Work Step by Step
$\mu$ =1262, $\sigma$ = 118
i ) z- score corresponding to lower 0.5 percentile (area of 0.005) = -2.5758
ii) z-score corresponding to upper 0.5 percentile = 2.5758
iii) Sub $\sigma = 118 , \mu= 1262, z = 2.5758, z = -2.5758 $ into z = $\frac{x - \mu}{\sigma}$ and solve for x:
z = $\frac{x - \mu}{\sigma}$
$x$ = $\mu$ + $z\sigma$
$x$ = $1262 + (-2.5758 \times 118)$
$x$ = $958.0556$ chips
$x$ $\approx$ 958 chips
z = $\frac{x - \mu}{\sigma}$
$x$ = $\mu$ + $z\sigma$
$x$ = $1262 + (2.5758 \times 118)$
$x$ = $1565.9444$ words
$x$ $\approx$ 1566 chips
