### Question 30**Prompt:**If the average weight for cats is 12 pounds and the standard deviation is 2 pounds, 68 percent of all cats will have a weight between __________ pounds and __________ pounds.**Explanation:**To find the range that includes 68 percent of all cats' weights, you need to calculate one standard deviation above and below the mean. - The mean weight of the cats is 12 pounds.- The standard deviation is 2 pounds.68% of values in a normal distribution fall within one standard deviation of the mean.1. Calculate the lower bound:\[ \text{Lower bound} = \text{Mean} - \text{Standard Deviation} \]\[ \text{Lower bound} = 12 \text{ pounds} - 2 \text{ pounds} \]\[ \text{Lower bound} = 10 \text{ pounds} \]2. Calculate the upper bound:\[ \text{Upper bound} = \text{Mean} + \text{Standard Deviation} \]\[ \text{Upper bound} = 12 \text{ pounds} + 2 \text{ pounds} \]\[ \text{Upper bound} = 14 \text{ pounds} \]Therefore, 68 percent of all cats will have a weight between 10 pounds and 14 pounds.### Question 31**Prompt:**Thirteen percent of adults are left-handed. In a group of 30 adults, find the probability of finding at least one lefty.1. **Option 1:** - No answer text provided.2. **Option 2:** - No answer text provided.3. **Option 3:** - 0.7984. **Option 4:** - 0.069**Explanation:**To find the probability of at least one left-handed person in a group of 30 adults, we can use the complement rule. Firstly, determine the probability of not finding any left-handed individuals and then subtract this from 1.- Probability of an adult being left-handed \(P(L) = 0.13\)- Probability of an adult not being left-handed \(P(R) = 1 - P(L) = 1 - 0.13 = 0.87\)The probability that all 30 adults are not left-handed:\[ P(\text{all **Educational Content Transcription**---### Question 30If the average weight for cats is 12 pounds and the standard deviation is 2 pounds, 68 percent of all cats will have a weight between _________ pounds and _________ pounds.### Question 31Thirteen percent of adults are left-handed. In a group of 30 adults, find the probability of finding at least one lefty.- [ ] No answer text provided.- [ ] No answer text provided.- [ ] 0.798- [ ] 0.069- [ ] 0.973- [ ] 0.985---**Explanation:**#### Question 30This question involves understanding the normal distribution. According to the empirical rule (68-95-99.7 rule), 68% of data within a normal distribution falls within one standard deviation of the mean. Here, the mean weight for cats is 12 pounds and the standard deviation is 2 pounds. Therefore, 68% of all cats will have a weight within one standard deviation of the mean, which means between:\[ 12 - 2 \text{ pounds and } 12 + 2 \text{ pounds} \]\[ 10 \text{ pounds and } 14 \text{ pounds} \]So, the answer should be:\[ 10 \text{ pounds and } 14 \text{ pounds} \]#### Question 31This question requires the use of binomial probability to find the likelihood of at least one event occurring. The percentage of left-handed adults is given as 13%. Given a group of 30 adults, the probability of finding at least one left-handed person can be calculated as follows:First, calculate the probability of not finding a lefty:\[ P(\text{not left-handed}) = 1 - P(\text{left-handed}) = 1 - 0.13 = 0.87 \]Then, find the probability of not finding a left-handed person in 30 people:\[ P(\text{no left-handed in 30}) = 0.87^{30} \]Finally, the probability of finding at least one left-handed person is:\[ P(\text{at least one}) = 1 - P(\text{no left-handed}) = 1 - 0.87^{30} \]This simplifies to approximately:\[ P(\text{at least

Name: ### Question 30**Prompt:**If the average weight for cats is 12 pounds
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Description: ### Question 30**Prompt:**If the average weight for cats is 12 pounds and the standard deviation is 2 pounds, 68 percent of all cats will have a weight between __________ pounds and __________ pounds.**Explanation:**To find the range that includes 6