The lengths of a particular snake are approximately normally distributed with a given mean [tex]$\mu=15 in.$[/tex] and standard deviation [tex]$\sigma=0.8 in.$[/tex]. What percentage of the snakes are longer than 16.6 in.?
A. 0.3%
B. 2.5%
C. 3.5%
D. 5%
Answer (2)
The percentage of snakes longer than 16.6 inches is approximately 2.5%. This was determined by calculating the Z-score and finding the corresponding probability. Final calculations confirm that the answer aligns with one of the provided options.
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Calculate the Z-score: Z = 0.8 16.6 − 15 = 2 .
Find the probability 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228"> P ( Z > 2 ) = 1 − P ( Z < 2 ) = 1 − 0.9772 = 0.0228 .
Convert the probability to a percentage: 0.0228 × 100 = 2.28% .
The percentage of snakes longer than 16.6 inches is approximately 2.5% .
Explanation
Understand the problem and provided data We are given that the lengths of snakes are normally distributed with a mean μ = 15 inches and a standard deviation σ = 0.8 inches. We want to find the percentage of snakes that are longer than 16.6 inches.
Calculate the Z-score Let X be the length of a snake. Then X follows a normal distribution with mean μ = 15 and standard deviation σ = 0.8 . We want to find 16.6)"> P ( X > 16.6 ) . To do this, we first calculate the Z-score for X = 16.6 using the formula: Z = σ X − μ Substituting the given values, we get: Z = 0.8 16.6 − 15 = 0.8 1.6 = 2
Find the probability using Z-score Now we need to find the probability that a snake is longer than 16.6 inches, which is equivalent to finding the area to the right of the calculated Z-score: 16.6) = P(Z > 2)"> P ( X > 16.6 ) = P ( Z > 2 ) Using a Z-table or calculator, we find the value of 2)"> P ( Z > 2 ) . The value of P ( Z < 2 ) is approximately 0.9772. Therefore, 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228"> P ( Z > 2 ) = 1 − P ( Z < 2 ) = 1 − 0.9772 = 0.0228
Convert to percentage and conclude To convert the probability to a percentage, we multiply by 100: 0.0228 × 100 = 2.28% Rounding to one decimal place, we get 2.3%. The closest option is 2.5%.
Examples
Understanding normal distributions and calculating probabilities is crucial in various fields. For instance, in manufacturing, it helps determine the percentage of products falling within acceptable quality ranges. In finance, it's used to assess the risk associated with investments. In healthcare, it can help determine the percentage of patients responding positively to a treatment.
