Question

shalise competed in a jigsaw puzzle competition where participants are timed on how long they take to complete puzzles of various sizes. shalise completed a small puzzle in 75 minutes and a large jigsaw puzzle in 140 minutes. for all participants, the distribution of completion time for the small puzzle was approximately normal with mean 60 minutes and standard deviation 15 minutes. the distribution of completion time for the large puzzle was approximately normal with mean 180 minutes and standard deviation 40 minutes. approximately what percent of the participants had finishing times greater than shalise’s for each puzzle?
a 16% on the small puzzle and 16% on the large puzzle
b 16% on the small puzzle and 84% on the large puzzle
c 32% on the small puzzle and 68% on the large puzzle
d 84% on the small puzzle and 84% on the large puzzle
e 84% on the small puzzle and 16% on the large puzzle

Explanation:

Step1: Analyze small puzzle (z - score)

For the small puzzle, mean $\mu = 60$, standard deviation $\sigma = 15$, Shalise's time $x = 75$.
Z - score formula: $z=\frac{x - \mu}{\sigma}$
$z=\frac{75 - 60}{15}=\frac{15}{15}=1$

In a normal distribution, about 68% of data is within $\mu\pm\sigma$, so above $z = 1$ (greater than Shalise's time) is $\frac{100\% - 68\%}{2}=16\%$? Wait, no: The area to the right of $z = 1$ in standard normal distribution is approximately $1 - 0.84 = 0.16$? Wait, no, actually, the cumulative probability for $z = 1$ is about 0.84 (from standard normal table), so the proportion above $z = 1$ is $1 - 0.84 = 0.16$? Wait, no, wait: Wait, mean is 60, $\sigma = 15$. 60 + 15 = 75, so Shalise's time is 1 standard deviation above the mean. In normal distribution, the percentage of data above $\mu+\sigma$: since 68% is within $\mu\pm\sigma$, the remaining 32% is outside, so 16% above $\mu+\sigma$ and 16% below $\mu - \sigma$. So for small puzzle, percentage of participants with time > 75 is 16%? Wait, no, wait: Wait, the cumulative probability for $z = 1$ is about 0.84 (from standard normal distribution: $P(Z\leq1)\approx0.84$), so $P(Z > 1)=1 - 0.84 = 0.16$, so 16% for small puzzle? Wait, no, wait, maybe I mixed up. Wait, Shalise's time is 75, which is $\mu+\sigma$ (60 + 15). So the area to the right of 75 (greater than 75) is the proportion of participants with time > 75. In normal distribution, the area to the right of $\mu+\sigma$: since the total area is 1, and the area between $\mu - \sigma$ and $\mu+\sigma$ is 0.68, so the area above $\mu+\sigma$ is $\frac{1 - 0.68}{2}=0.16$, so 16%? Wait, no, that's if we consider both tails. Wait, no, the standard normal distribution: $z = 1$ corresponds to the 84th percentile (since $P(Z\leq1)\approx0.84$), so the percentage of people with time greater than 75 is $100\% - 84\% = 16\%$. So small puzzle: 16%.

Step2: Analyze large puzzle (z - score)

For the large puzzle, mean $\mu = 180$, standard deviation $\sigma = 40$, Shalise's time $x = 140$.
Z - score: $z=\frac{140 - 180}{40}=\frac{-40}{40}=-1$

Cumulative probability for $z = - 1$ is $P(Z\leq - 1)\approx0.16$ (from standard normal table). So the proportion of participants with time > 140 is $1 - 0.16 = 0.84$, so 84%.

Answer:

B. 16% on the small puzzle and 84% on the large puzzle