The lengths of salamanders have a normal distribution with mean 15cm, and standard deviation 2cm. What length of salamander would place it at the 80th percentile of salamander lengths?

Answers

Answer 1

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

Answer 2

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


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