4.4.3 Test (CST) Continuous Probability Distributions
PLEASE HELP!!!
The head-to-tail length of a species of fish is normally distributed, with a mean of 17.4 cm and a standard deviation of 3.2 cm. One fish is 15.3 cm long. What is the z-score of the length of this fish? Round your answer to two decimal places.
A. -0.86
B. -0.66
C. 0.86
D. 0.66

Answer:

B

Step-by-step explanation:

We have that for a normal distribution, the value of z can be calculated using the following formula:

z = (x - m) / sd

where x is the value to evaluate, m the mean and sd the standard deviation, we have those values x = 15.3, m = 17.4 and sd = 3.2

if we replace we have:

z = (15.3 - 17.4) /3.2

z = -0.65625

if we round to 2 decimal places, it would be -0.66, that is, the answer is B.

A Z-score helps us to understand how far the data is from the mean. The z-score of the given length of this fish is -0.66. The correct option is B.

A Z-score helps us to understand how far is the data from the mean. It is a measure of how many times the data is above or below the mean. It is given by the formula,

z-score = (x-μ)/σ

Where Z is the Z-score,

X is the data point,

μ is the mean and σ is the standard variable.

Given that the length of a species of fish is normally distributed, with a mean(μ) and the standard deviation(σ) is 17.4 cm and 3.2 cm. Therefore, the z-score of the length of this fish is,

z-score = (x-μ)/σ

            = (15.3 - 17.4) / 3.2

            = -2.1/3.2

            = -0.65625

            = -0.66

Hence, the z-score of the given length of this fish is -0.66.

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