Question

question 6 of 6 the weights of laboratory cockroaches can be modeled with a normal distribution having mean 80 grams and standard deviation 2 grams. the following figure shows the normal curve for this distribution of weight. about what proportion of the cockroaches will have weights greater than 83 grams? 0.0668 0.0772 0.0228 0.1587 0.9332 incorrect answer normal curve image below copyright © bfw publishers

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 83$, $\mu=80$, and $\sigma = 2$ (assuming the standard deviation is 2, since the mean is 80 and we can infer from the normal distribution context).
So, $z=\frac{83 - 80}{2}=\frac{3}{2}=1.5$.

Step2: Find the proportion using z - table

We want to find $P(X>83)$, which is equivalent to $P(Z > 1.5)$ in the standard normal distribution.
We know that $P(Z\leq1.5)$ can be found from the standard normal table. Looking up $z = 1.5$ in the standard normal table, we get $P(Z\leq1.5)=0.9332$.
Then, $P(Z > 1.5)=1 - P(Z\leq1.5)=1 - 0.9332 = 0.0668$. Wait, no, wait. Wait, maybe I made a mistake. Wait, if the standard deviation is 2, mean is 80. Wait, maybe the standard deviation is 2? Wait, no, maybe the normal distribution has mean 80 and standard deviation 2? Wait, but let's re - check. Wait, the options are 0.0668, 0.0772, 0.0228, 0.1587, 0.9332. Wait, maybe the standard deviation is 2? Wait, 83 is 1.5 standard deviations above the mean (since $83-80 = 3$, and if $\sigma=2$, then $z = 1.5$). But the area to the right of $z = 1.5$ is $1 - 0.9332=0.0668$? But wait, maybe the standard deviation is 1? No, that can't be. Wait, maybe I misread the standard deviation. Wait, the problem says "standard deviation 2 grams"? Wait, the original problem says "standard deviation 2 grams"? Wait, the user's image: "The weights of laboratory cockroaches can be modeled with a normal distribution having mean 80 grams and standard deviation 2 grams". Oh, right, standard deviation $\sigma = 2$. So $z=\frac{83 - 80}{2}=1.5$. Then $P(X>83)=P(Z > 1.5)$. From the standard normal table, $P(Z\leq1.5)=0.9332$, so $P(Z > 1.5)=1 - 0.9332 = 0.0668$? But wait, the incorrect answer was 0.1587. Wait, maybe I made a mistake. Wait, no, wait, maybe the standard deviation is 1? No, the mean is 80, standard deviation 2. Wait, 83 is 1.5 sigma above mean. Wait, but let's check the empirical rule. For a normal distribution, about 68% within $\mu\pm\sigma$, 95% within $\mu\pm2\sigma$, 99.7% within $\mu\pm3\sigma$. $\mu = 80$, $\sigma=2$. So $\mu+\sigma=82$, $\mu + 1.5\sigma=80 + 3=83$. The area to the right of $\mu+1.5\sigma$: the area between $\mu$ and $\mu + 1.5\sigma$ is about 43.32% (since up to $\mu+1\sigma$ is 34%, up to $\mu + 1.5\sigma$ is 43.32%), so the area to the right is $50\% - 43.32\%=6.68\% = 0.0668$. But wait, the options have 0.0668 as an option. But the incorrect answer was 0.1587. Wait, maybe the standard deviation is 1? No, the problem says standard deviation 2. Wait, maybe I misread the mean. Wait, mean is 80, standard deviation 2. So $z=\frac{83 - 80}{2}=1.5$. So the proportion is 0.0668. But let's check the z - table again. The z - score of 1.5 corresponds to a cumulative probability of 0.9332, so the area to the right is $1 - 0.9332 = 0.0668$. So the correct proportion is 0.0668.

Answer:

0.0668