QCAA Mathematical Methods Continuous random variables and the normal distribution

This range falls between 2 and 4 standard deviations below the mean ($z = -4$ to $z = -2$). This represents the far left tail of the curve, which contains a negligible percentage of the population.

This range covers the area from 2 standard deviations below the mean up to the mean ($z = -2$ to $z = 0$). While substantial, it captures less area than an interval of the same width centered directly on the peak of the distribution.

This range corresponds to exactly one standard deviation below and above the mean ($15 \pm 2$ cm). Because the normal distribution is symmetric and peaks at the mean, the interval centered on the mean contains the largest proportion of data (approximately 68%).

This range falls between 1 and 3 standard deviations above the mean ($z = 1$ to $z = 3$). This represents the tapering right tail of the distribution, which contains significantly fewer teenagers than the central range.

This value represents the probability for the interval $10 \leq t < 15$ only ($30/200 = 0.15$). You must also include the interval $15 \leq t < 20$.

The total number of services is $10+20+30+40+100=200$. The number of services between 10 and 20 minutes is $30+40=70$. The probability is $\frac{70}{200} = 0.35$.

This option likely confuses the frequency count (70) with the probability, or assumes a total frequency of 100. You must divide the specific frequency (70) by the total frequency (200).

This is the probability that the service took at least 10 minutes ($t \geq 10$), calculated as $\frac{30+40+100}{200} = 0.85$. The question excludes times of 20 minutes or more.