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 shalises 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

Explanation:

Step1: Calculate z - score for small puzzle

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean and $\sigma$ is the standard deviation. For the small puzzle, $x = 75$, $\mu=60$, $\sigma = 15$. So $z=\frac{75 - 60}{15}=\frac{15}{15}=1$.

Step2: Find the proportion for small puzzle

Using the standard normal distribution table, the proportion of values less than $z = 1$ is approximately $0.8413$. So the proportion of values greater than $z = 1$ is $1-0.8413\approx0.16$ or $16\%$.

Step3: Calculate z - score for large puzzle

For the large puzzle, $x = 140$, $\mu = 180$, $\sigma=40$. So $z=\frac{140 - 180}{40}=\frac{- 40}{40}=-1$.

Step4: Find the proportion for large puzzle

Using the standard normal distribution table, the proportion of values less than $z=-1$ is approximately $0.1587$. So the proportion of values greater than $z=-1$ is $1 - 0.1587\approx0.84$ or $84\%$.

Answer:

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