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Debora
1 month ago
11

The lengths of a particular snake are approximately normally distributed with a given mean mc025-1.jpg = 15 in. and standard dev

iation mc025-2.jpg = 0.8 in. What percentage of the snakes are longer than 16.6 in.?
0.3%
2.5%
3.5%
5%

Mathematics
2 answers:
lawyer [12.5K]1 month ago
6 0
The answer to the previously mentioned question is: "3.5%" The lengths of a particular snake follow an approximately normal distribution with a mean of mc025-1.jpg = 15 inches and a standard deviation of mc025-2.jpg = 0.8 inches. The percentage of snakes that exceed 16.6 inches is 3.5%

PIT_PIT [12.4K]1 month ago
6 0

Answer:

2.5% of the snakes exceed the length of 16.6 inches

Detailed explanation:

The Z score is calculated as follows:

Z=\dfrac{X-\mu}{\sigma}

Wherein:

X = raw score = 16.6

μ = mean = 15

σ = standard deviation = 0.8

<pconsequently>

Z=\dfrac{16.6-15}{0.8}=2

The Z-score signifies how far the data diverges from the mean in terms of standard deviations. Hence, the value 16.6 lies 2 standard deviations above the mean.

About 95% of all data falls within two standard deviations from the mean. (spanning from mean-2 times the standard deviation to mean+2 times the standard deviation)

Thus, 5% of the total data is outside this range. To determine how many snakes exceed 16.6 inches, we must identify the area to the right of this length.

This remaining 5% is evenly split into two parts.

Consequently, \dfrac{5}{2}=2.5\% of the snakes surpass 16.6 inches

</pconsequently>
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