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
The length of salamander that would place it at the 80th percentile of salamander lengths is 16.684 cm and this can be determined by using the formula of z-score.
Given :
The lengths of salamanders have a normal distribution with a mean of 15 cm, and a standard deviation of 2 cm.
The following steps can be used in order to determine the length of the salamander:
Step 1 - The formula of z-score can be used in order to determine the length of the salamander.
[tex]z = \dfrac{x-\mu}{\sigma}[/tex] --- (1)
Step 2 - For the value of [tex]P(z\leq x)=0.8[/tex] the value of 'z' is 0.842.
Step 3 - Now, substitute the values of z, [tex]\mu[/tex], and [tex]\sigma[/tex] in the expression (1).
[tex]0.842 = \dfrac{x-15}{2}[/tex]
Step 4 - Simplify the above expression.
[tex]x = 15 + 1.684\\x = 16.684[/tex]
So, the length of salamander that would place it at the 80th percentile of salamander lengths is 16.684 cm.
For more information, refer to the link given below:
https://brainly.com/question/13299273
