The lengths of salamanders have a normal distribution with mean 15cm, and standard deviation 2cm. What length of salamander woul
d place it at the 80th percentile of salamander lengths?
1 answer:
Answer:
16.68 cm
Step-by-step explanation:
Given that :
Mean , m = 15
Standard deviation, s = 2
Length of salamander that would Place it at 80% of salamander length :
P(Z ≤ x) = 0.8
Zscore equivalent to 0.8 = 0.842
Using the relation :
Zscore = (x - m) /s
0.842 = (x - 15) / 2
1.684 = x - 15
Add 15 to both sides
1.684 + 15 = x - 15 + 15
16.684 = x
Hence, x = 16.68 cm
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