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Sagot :
To determine the percentage of snakes that are longer than 16.6 inches, we need to follow the steps for solving problems involving the normal distribution. Here's the detailed, step-by-step solution:
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]) = 15 inches
- Standard deviation ([tex]\(\sigma\)[/tex]) = 0.8 inches
- Threshold length for comparison = 16.6 inches
2. Convert the length threshold to a z-score:
A z-score tells us how many standard deviations away from the mean a particular value lies. The formula for the z-score ([tex]\(z\)[/tex]) is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value of interest (here, 16.6 inches).
Plugging in the values:
[tex]\[ z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2 \][/tex]
So, the z-score for 16.6 inches is 2.
3. Find the cumulative probability associated with the calculated z-score:
We look up the z-score of 2 in the standard normal distribution table (or use a cumulative distribution function calculator) to find the probability that a randomly selected value from a normal distribution is less than or equal to the z-score.
The cumulative probability (also known as the cumulative distribution function, or CDF) for [tex]\(z = 2\)[/tex] is approximately 0.9772. This means that approximately 97.72% of the observations are less than or equal to 16.6 inches.
4. Calculate the probability of the snake being longer than 16.6 inches:
To find the percentage of snakes longer than 16.6 inches, we need to find the complementary probability (i.e., the probability that a snake's length is more than the threshold).
This is given by:
[tex]\[ P(X > 16.6) = 1 - P(X \leq 16.6) = 1 - 0.9772 = 0.0228 \][/tex]
Thus, the probability that a snake is longer than 16.6 inches is 0.0228 or 2.28%.
5. Convert the probability to a percentage:
[tex]\[ 0.0228 \times 100 = 2.28\% \][/tex]
Therefore, the percentage of snakes that are longer than 16.6 inches is approximately [tex]\(2.3\%\)[/tex].
Given the provided options, the correct answer is:
[tex]\[ 2.5\% \][/tex]
(Note: While the accurate value we calculated is 2.28%, the closest option is 2.5%.)
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]) = 15 inches
- Standard deviation ([tex]\(\sigma\)[/tex]) = 0.8 inches
- Threshold length for comparison = 16.6 inches
2. Convert the length threshold to a z-score:
A z-score tells us how many standard deviations away from the mean a particular value lies. The formula for the z-score ([tex]\(z\)[/tex]) is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X\)[/tex] is the value of interest (here, 16.6 inches).
Plugging in the values:
[tex]\[ z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2 \][/tex]
So, the z-score for 16.6 inches is 2.
3. Find the cumulative probability associated with the calculated z-score:
We look up the z-score of 2 in the standard normal distribution table (or use a cumulative distribution function calculator) to find the probability that a randomly selected value from a normal distribution is less than or equal to the z-score.
The cumulative probability (also known as the cumulative distribution function, or CDF) for [tex]\(z = 2\)[/tex] is approximately 0.9772. This means that approximately 97.72% of the observations are less than or equal to 16.6 inches.
4. Calculate the probability of the snake being longer than 16.6 inches:
To find the percentage of snakes longer than 16.6 inches, we need to find the complementary probability (i.e., the probability that a snake's length is more than the threshold).
This is given by:
[tex]\[ P(X > 16.6) = 1 - P(X \leq 16.6) = 1 - 0.9772 = 0.0228 \][/tex]
Thus, the probability that a snake is longer than 16.6 inches is 0.0228 or 2.28%.
5. Convert the probability to a percentage:
[tex]\[ 0.0228 \times 100 = 2.28\% \][/tex]
Therefore, the percentage of snakes that are longer than 16.6 inches is approximately [tex]\(2.3\%\)[/tex].
Given the provided options, the correct answer is:
[tex]\[ 2.5\% \][/tex]
(Note: While the accurate value we calculated is 2.28%, the closest option is 2.5%.)
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