Sample size is 275: Generally we require sample size > 30 for the normal distribution, so this is enough.

Standard deviation of a proportion p is given by sqrt(p(1-p))

sd = sqrt(0.2 x 0.8)

= 0.4

So the standard error for a sample mean is given by sd/sqrt(n) where n is the sample size:

SEM = sd/sqrt(n)

= 0.4/sqrt(275)

= 0.0241

So the probability that fewer than 15% involve driver distraction in our sample is the normal cdf at .15 with mean = .2 and sd = SEM = 0.0241:

Norm.cdf (sample mean, population mean, standard error)

= Norm.cdf (0.15, 0.2, 0.0241)

= 0.0191

And the probability that more than 25% involve driver distraction is the same (it is also 5% from the population mean, and the normal distribution is symmetric around the mean).

So the probability that between 15% and 25% involve driver distraction will be: