Suppose the average length of a fish caught in Lake Springfield is 10.6 in with standard deviation 4.1 in. Assuming lengths of fish caught in Lake Springfield are normally distributed, find the probability that a fish caught there will be longer than 18 in.

0.035547

Step-by-step explanation:

We would be using z score formula to solve for this question

The formula for calculating a z-score is is

z = (x-μ)/σ

where x is the raw score = 18 in

μ is the population mean = 10.6 in

σ is the population standard deviation = 4.1 in

z score = 18 - 10.6/ 4.1

z score = 7.4 /4.1

z score = 1.80488

Using the normal distribution table to find the probability for the z score of 1.80488

Probability value from Z-Table:

P(z = 1.80488) = 0.96445

The probability that a fish caught there will be longer than 18 in is calculated as:

P(x>18) = 1 - P(z = 1.80488)

= 1 - 0.96445

= 0.035547

side xy = 10

sin(x) = 4/5

cos(x) = 3/5

tan(x) = 4/3

step-by-step explanation:

table 3 represents a direct variation function. : )