At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To determine the percentage of snakes that are longer than 16.6 inches, we can follow these detailed steps:
1. Understand the Parameters of the Distribution:
The lengths of the snakes are normally distributed with a mean \(\mu = 15\) inches and a standard deviation \(\sigma = 0.8\) inches.
2. Calculate the Z-Score:
The Z-score helps us determine how many standard deviations a particular value (in this case, 16.6 inches) is from the mean.
The formula for the Z-score is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where \(X\) is the value we are interested in (16.6 inches).
Plugging in the values:
[tex]\[ Z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2 \][/tex]
3. Determine the Cumulative Probability:
The Z-score tells us the number of standard deviations above the mean the value lies. We use the standard normal distribution to find the cumulative probability up to this Z-score.
The cumulative probability for \(Z = 2\) from standard normal distribution tables or software is approximately 0.9772. This represents the probability that a snake is shorter than or equal to 16.6 inches.
4. Calculate the Complement to Find the Required Probability:
We are interested in the percentage of snakes that are longer than 16.6 inches. This is the complement of the cumulative probability we found.
[tex]\[ P(X > 16.6) = 1 - P(X \leq 16.6) \][/tex]
So, substituting in the cumulative probability:
[tex]\[ P(X > 16.6) = 1 - 0.9772 = 0.0228 \][/tex]
5. Convert the Probability to a Percentage:
To convert this probability to a percentage, we multiply by 100:
[tex]\[ 0.0228 \times 100 = 2.28\% \][/tex]
Therefore, approximately 2.28% of the snakes are longer than 16.6 inches. The closest answer choice to this percentage is 2.5%. Hence, the correct answer is:
[tex]\[ \boxed{2.5\%} \][/tex]
1. Understand the Parameters of the Distribution:
The lengths of the snakes are normally distributed with a mean \(\mu = 15\) inches and a standard deviation \(\sigma = 0.8\) inches.
2. Calculate the Z-Score:
The Z-score helps us determine how many standard deviations a particular value (in this case, 16.6 inches) is from the mean.
The formula for the Z-score is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where \(X\) is the value we are interested in (16.6 inches).
Plugging in the values:
[tex]\[ Z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2 \][/tex]
3. Determine the Cumulative Probability:
The Z-score tells us the number of standard deviations above the mean the value lies. We use the standard normal distribution to find the cumulative probability up to this Z-score.
The cumulative probability for \(Z = 2\) from standard normal distribution tables or software is approximately 0.9772. This represents the probability that a snake is shorter than or equal to 16.6 inches.
4. Calculate the Complement to Find the Required Probability:
We are interested in the percentage of snakes that are longer than 16.6 inches. This is the complement of the cumulative probability we found.
[tex]\[ P(X > 16.6) = 1 - P(X \leq 16.6) \][/tex]
So, substituting in the cumulative probability:
[tex]\[ P(X > 16.6) = 1 - 0.9772 = 0.0228 \][/tex]
5. Convert the Probability to a Percentage:
To convert this probability to a percentage, we multiply by 100:
[tex]\[ 0.0228 \times 100 = 2.28\% \][/tex]
Therefore, approximately 2.28% of the snakes are longer than 16.6 inches. The closest answer choice to this percentage is 2.5%. Hence, the correct answer is:
[tex]\[ \boxed{2.5\%} \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.