The lengths of a particular snake are approximately normally distributed with a given mean [tex]\mu = 15[/tex] in. and standard deviation [tex]\sigma = 0.8[/tex] in. What percentage of the snakes are longer than 16.6 in.?

A. [tex]0.3 \%[/tex]
B. [tex]2.5 \%[/tex]
C. [tex]3.5 \%[/tex]
D. [tex]5 \%[/tex]



Answer :

To determine the percentage of snakes that are longer than 16.6 inches, we can follow these steps:

1. Identify the Given Values:
- Mean length of snakes (\(\mu\)): 15 inches
- Standard deviation of lengths (\(\sigma\)): 0.8 inches
- Length threshold to consider: 16.6 inches

2. Calculate the Z-Score:
The Z-score is a measure of how many standard deviations an element is from the mean. It can be calculated using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where \(X\) is the value of the threshold length (16.6 inches).

Plugging in the values, we get:
[tex]\[ Z = \frac{16.6 - 15}{0.8} = \frac{1.6}{0.8} = 2 \][/tex]

3. Determine the Cumulative Probability:
The cumulative probability associated with a Z-score tells us the proportion of values below that Z-score in a standard normal distribution. For a Z-score of 2, the cumulative probability is approximately 0.9772 (or 97.72%).

4. Calculate the Percentage Above the Threshold:
To find the percentage of snakes longer than 16.6 inches, we need to consider the proportion of the distribution above this Z-score. Since the total area under the normal distribution curve is 1 (or 100%), the area above the Z-score of 2 is:
[tex]\[ 1 - \text{Cumulative Probability} = 1 - 0.9772 = 0.0228 \][/tex]

Converting this to a percentage:
[tex]\[ 0.0228 \times 100\% = 2.28\% \][/tex]

Therefore, the percentage of snakes longer than 16.6 inches is approximately 2.28%. Among the given options, the closest value is:

[tex]\[ \boxed{2.5\%} \][/tex]