3. the price of a gallon of regular gasoline at 75 gas stations across the state is normally distributed with a mean of $2.05 and a standard deviation of 4¢.
a) what percent of gas stations sell a gallon of regular gas for less than $1.97?
b) what percent of gas stations sell a gallon of regular gas for at least $2.17?
c) what is the probability that a gas station sells a gallon of regular gas for less than $1.97 or greater than $2.05?
d) about how many stations sell a gallon of regular gas for no more than $2.01?
4. mrs. fuller recently tested her 120 keyboarding students to see how many words per minute they can type. the results were normally distributed with a mean of 45 and a standard deviation of 6.
a) about how many students can type at least 39 words per minute?
b) about how many students can type within one standard deviation of the mean?
c) students need to be in the top 2% in order to be eligible for the national typing competition. if carla can type 56 wpm, is she eligible?

Question

Accepted Answer

### Problem 3a # Explanation: ## Step1: Identify the normal distribution The data is normally distributed with mean $\mu = 2.05$ and standard deviation $\sigma = 0.04$ (since 4¢ = $0.04$). We need to find the percentage of data less than $1.97$. ## Step2: Calculate the z - score The z - score formula is $z=\frac{x - \mu}{\sigma}$. For $x = 1.97$, $z=\frac{1.97 - 2.05}{0.04}=\frac{- 0.08}{0.04}=- 2$. ## Step3: Use the empirical rule For a normal distribution, the percentage of data less than $\mu-2\sigma$ (z - score=- 2) is the sum of the percentages in the left - most two tails. From the empirical rule, the percentage less than $\mu - 2\sigma$ is $2.2\%+0.1\% = 2.3\%$. # Answer: $2.3\%$ ### Problem 3b # Explanation: ## Step1: Identify the normal distribution Mean $\mu = 2.05$, standard deviation $\sigma = 0.04$. We need to find the percentage of data at least $2.17$ (i.e., $x\geq2.17$). ## Step2: Calculate the z - score For $x = 2.17$, $z=\frac{2.17 - 2.05}{0.04}=\frac{0.12}{0.04}=3$. ## Step3: Use the empirical rule For a normal distribution, the percentage of data greater than or equal to $\mu + 3\sigma$ (z - score = 3) is $0.1\%$. # Answer: $0.1\%$ ### Problem 3c # Explanation: ## Step1: Find the percentage less than $1.97$ From part 3a, the percentage of data less than $1.97$ is $2.3\%$. ## Step2: Find the percentage greater than $2.05$ The mean $\mu = 2.05$. In a normal distribution, the percentage of data greater than the mean is $50\%$ (since the normal distribution is symmetric about the mean). ## Step3: Add the two percentages The probability of a gas station selling gas for less than $1.97$ or greater than $2.05$ is $2.3\%+50\%=52.3\%$. # Answer: $52.3\%$ ### Problem 3d # Explanation: ## Step1: Identify the normal distribution Mean $\mu = 2.05$, standard deviation $\sigma = 0.04$. We need to find the number of stations with price no more than $2.01$ (i.e., $x\leq2.01$). ## Step2: Calculate the z - score For $x = 2.01$, $z=\frac{2.01 - 2.05}{0.04}=\frac{-0.04}{0.04}=- 1$. ## Step3: Use the empirical rule The percentage of data less than $\mu-\sigma$ (z - score=-1) is $13.6\%+2.2\%+0.1\% = 15.9\%$? Wait, no. Wait, the empirical rule: the percentage of data between $\mu-\sigma$ and $\mu+\sigma$ is $68\%$, between $\mu - 2\sigma$ and $\mu+2\sigma$ is $95\%$, between $\mu-3\sigma$ and $\mu + 3\sigma$ is $99.7\%$. The percentage less than $\mu-\sigma$ is $\frac{100\% - 68\%}{2}=16\%$ (approx). Wait, from the given graph, the percentage less than $2.01$ (which is $\mu-\sigma$) is $13.6\%+2.2\%+0.1\%=15.9\%\approx16\%$? Wait, the graph has $0.1\%$, $2.2\%$, $13.6\%$ on the left of the mean. So the percentage less than $2.01$ ($\mu-\sigma$) is $0.1\%+2.2\%+13.6\% = 15.9\%\approx16\%$. But the total number of stations is 75. ## Step4: Calculate the number of stations Number of stations $=75 imes0.159\approx12$ (or using the graph's values: the percentage less than $2.01$ is $0.1 + 2.2+13.6 = 15.9\%$, $75 imes0.159\approx12$). Wait, but let's re - check. The mean is $2.05$, $\sigma = 0.04$. $2.01=2.05 - 0.04=\mu-\sigma$. The percentage of data less than $\mu-\sigma$ is $\frac{1 - 0.68}{2}=0.16 = 16\%$. So $75 imes0.16 = 12$. # Answer: Approximately 12 stations. ### Problem 4a # Explanation: ## Step1: Identify the normal distribution Mean $\mu = 45$, standard deviation $\sigma = 6$. We need to find the number of students who can type at least 39 words per minute (i.e., $x\geq39$). ## Step2: Calculate the z - score For $x = 39$, $z=\frac{39 - 45}{6}=\frac{-6}{6}=-1$. ## Step3: Use the empirical rule The percentage of data greater than or equal to $\mu-\sigma$ (z - score=-1) is $100\%-\frac{100\% - 68\%}{2}=84\%$ (since the percentage less than $\mu-\sigma$ is $\frac{100 - 68}{2}=16\%$, so the percentage greater than or equal to $\mu-\sigma$ is $100 - 16=84\%$). ## Step4: Calculate the number of students Number of students $=120 imes0.84 = 100.8\approx101$ (or more accurately, using the empirical rule: the percentage of data $\geq\mu-\sigma$ is $68\%+16\% = 84\%$ (since between $\mu-\sigma$ and $\mu+\sigma$ is $68\%$, and above $\mu+\sigma$ is $16\%$, wait no: the normal distribution is symmetric. The percentage less than $\mu-\sigma$ is $16\%$, so the percentage greater than or equal to $\mu-\sigma$ is $100 - 16 = 84\%$). So $120 imes0.84 = 100.8\approx101$. # Answer: Approximately 101 students. ### Problem 4b # Explanation: ## Step1: Recall the empirical rule For a normal distribution, the percentage of data within one standard deviation of the mean ($\mu-\sigma$ to $\mu+\sigma$) is $68\%$. ## Step2: Calculate the number of students Number of students $=120 imes0.68 = 81.6\approx82$. # Answer: Approximately 82 students. ### Problem 4c # Explanation: ## Step1: Calculate the z - score for Carla's typing speed Carla's speed $x = 56$. Mean $\mu = 45$, standard deviation $\sigma = 6$. The z - score is $z=\frac{56 - 45}{6}=\frac{11}{6}\approx1.83$. ## Step2: Determine the percentage of data above z = 1.83 The top $2\%$ of data corresponds to a z - score of approximately 2 (since for z = 2, the percentage above z = 2 is $2.2\%+0.1\%=2.3\%\approx2\%$). Since Carla's z - score is approximately $1.83<2$, the percentage of students with typing speed greater than 56 is more than $2\%$? Wait, no. Wait, the z - score of 2 corresponds to $\mu + 2\sigma=45 + 12 = 57$. Carla's speed is 56, which is less than 57. The percentage of data above z = 2 is about $2.3\%$, and the percentage above z = 1.83 is more than $2.3\%$? Wait, no. Let's use the standard normal table. The area to the left of z = 1.83 is approximately 0.9664, so the area to the right (percentage of students with speed > 56) is $1 - 0.9664 = 0.0336=3.36\%>2\%$. Wait, but the top $2\%$ is the area to the right of z such that the area is 0.02. The z - score for which the area to the right is 0.02 is approximately 2.05 (from standard normal table). Since Carla's z - score is 1.83 < 2.05, the percentage of students with speed greater than 56 is more than $2\%$, so she is eligible? Wait, no. Wait, if the top $2\%$ is the fastest $2\%$, we need to see if Carla's speed is in the top $2\%$. The mean is 45, $\sigma = 6$. $\mu+2\sigma=45 + 12 = 57$, $\mu+3\sigma=45 + 18 = 63$. Carla's speed is 56, which is less than 57. The percentage of data above $\mu + 2\sigma$ is about $2.3\%$, so the percentage above 56 (which is less than 57) is more than $2.3\%$, so it is more than $2\%$. Wait, but maybe the question is using the empirical rule. The top $2\%$ is approximately the area above $\mu + 2\sigma$ (since the area above $\mu+2\sigma$ is $2.2\%+0.1\% = 2.3\%\approx2\%$). Since 56 < 57 ($\mu + 2\sigma$), the percentage of students with speed greater than 56 is more than $2\%$, so she is eligible? Wait, no. Wait, if the top $2\%$ is the fastest $2\%$, we need to check if her speed is in the top $2\%$. The z - score of 56 is $\frac{56 - 45}{6}=\frac{11}{6}\approx1.83$. The area to the right of z = 1.83 is $1 - 0.9664 = 0.0336 = 3.36\%$, which is more than $2\%$, so she is eligible? Wait, no, if the top $2\%$ is required, and her percentage is $3.36\%$, which is more than $2\%$, that means she is in the top $3.36\%$, which is within the top $2\%$? No, $3.36\%>2\%$, so she is not in the top $2\%$? Wait, I made a mistake. The top $2\%$ means that only $2\%$ of the students have a higher speed than her. If the area to the right of her z - score is $3.36\%$, that means $3.36\%$ of the students have a higher speed, which is more than $2\%$, so she is not eligible? Wait, let's re - calculate the z - score. $x = 56$, $\mu = 45$, $\sigma = 6$. $z=\frac{56 - 45}{6}=\frac{11}{6}\approx1.83$. Looking at the standard normal table, the cumulative probability for z = 1.83 is 0.9664, so the probability that a student has a speed greater than 56 is $1 - 0.9664 = 0.0336$, or $3.36\%$. Since $3.36\%>2\%$, there are more than $2\%$ of students with a higher speed, so she is not eligible? Wait, but maybe the question is using the empirical rule where the top $2\%$ is above $\mu+2\sigma$ (which is 57). Since 56 < 57, she is not in the top $2\%$ (because the top $2\%$ is above 57). So the answer is no? Wait, no, let's check again. The mean is 45, $\sigma = 6$. $\mu+2\sigma=45 + 12 = 57$, $\mu+3\sigma=45+18 = 63$. The percentage of data above $\mu + 2\sigma$ is $2.2\%+0.1\%=2.3\%\approx2\%$. So if a student's speed is above 57, they are in the top $2\%$. Carla's speed is 56, which is below 57, so she is not in the top $2\%$, so she is not eligible. # Answer: No, she is not eligible.

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QUESTION IMAGE

Question

Explanation:

Problem 3a

Step1: Identify the normal distribution

Step2: Calculate the z - score

Step3: Use the empirical rule

Step1: Identify the normal distribution

Step2: Calculate the z - score

Step3: Use the empirical rule

Step1: Find the percentage less than $1.97$

Step2: Find the percentage greater than $2.05$

Step3: Add the two percentages

Answer:

Problem 3b