Question

A fish story: The mean length of one-year-old spotted flounder, in millimeters, is 127 with standard deviation of 22, and the mean length of two-year-old spotted flounder is 158 with a standard deviation of 23. The distribution of flounder lengths is approximately bell-shaped. Part 1 of 4 (a) Anna caught a one-year-old flounder that was 155 millimeters in length. What is the z-score for this length

Answer 1

Answer:

The z-score for this length is of 1.27.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

One-year-old flounder:

Mean of 127 with standard deviation of 22, which means that $\mu = 127, \sigma = 22$

Anna caught a one-year-old flounder that was 155 millimeters in length. What is the z-score for this length

This is Z when X = 155. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{155 - 127}{22}$

$Z = 1.27$

The z-score for this length is of 1.27.