Confidence Intervals · Superchallenge problem

A normal sampling distribution has standard error 1, so the usual 95% interval for $\mu$ is $\bar x \pm 1.96$, of width $3.92$. An analyst instead builds a 95% interval that puts 4% of the miss probability on the low side and 1% on the high side: $(\bar x - z_{0.99},\ \bar x + z_{0.96})$, where $\Phi(z_{0.99}) = 0.99$ and $\Phi(z_{0.96}) = 0.96$. By how much is this interval WIDER than the usual one? Round to two decimal places.