Question

I have 4929 songs in my computers music library. The songs have a mean duration of 243.5 seconds with a standard deviation of 111.93 seconds. One of the songs is 389 seconds long. What is its z-score?

Answer 1

Given:

Mean duration (μ) = 243.5 seconds

Standard deviation (σ) = 111.93 seconds

One of the songs is 389 seconds long (x)

To be able to determine the z-score, we will be using the following formula:

`\text{ z = }\frac{\text{ x - }\mu\text{ }}{\sigma}`

Where,

z = the z-score

x = the score

μ = the mean

σ = the standard deviation

We get,

`\text{ z = }\frac{\text{ x - }\mu\text{ }}{\sigma}`
`\text{ z = }\frac{\text{ 389 - 243.5 }}{\text{ 111.93}}`
`\text{ z = }\frac{\text{ 145.5 }}{\text{ 111.93}}`
`\text{ z = }1.2999195926\text{ }\approx\text{ 1.3}`

Therefore, the z-score is 1.3