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A radio station claims that the amount of advertising per hour of broadcast time has an average of 17 minutes and a standard deviation equal to 2.2 minutes. You listen to the radio station for 1 hour, at a randomly selected time, and carefully observe that the amount of advertising time is equal to 15 minutes. Calculate the z-score for this amount of advertising time.

Respuesta :

Answer:

-0.91

Step-by-step explanation:

The z score formula is given as:

z = (x-μ)/σ, where

x is the raw score = 15 minutes

μ is the population mean = 17 minutes

σ is the population standard deviation = 2.2 minutes

z = 15 - 17/2.2

z = -2/2.2

z = -0.90909

Approximately = -0.91

The z-score for this amount of advertising time is -0.91

Cetacea

z score is -0.91

The z score formula is given as:

z = (x-μ)/σ, where

  • x is the raw score = 15 minutes
  • μ is the population mean = 17 minutes
  • σ is the population standard deviation = 2.2 minutes

[tex]z=\frac{\left(15-17\right)}{2.2}\\\\z=-0.909[/tex]

Approximately = -0.91

The z-score for this amount of advertising time is -0.91.

Learn more about z-score: https://brainly.com/question/25638875

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