The length of country & western songs has mean 151 seconds and standard deviation 30 seconds. Determine the probability (as percent) that a random selection of 20 songs will have mean length of less than 149.75 seconds.

48.514%

Step-by-step explanation:

The formula for calculating a z-score is is z = (x -μ)/σ,

where x is the raw score = 149.75

μ is the sample mean = population mean = 151 seconds

σ is the sample standard deviation

= Population standard deviation/√n

Where n = 20 songs

Population standard deviation = 30

Hence, sample standard deviation

30/√20

= 6.7082039325

z = (x -μ)/σ

z = 149.75 - 150/6.7082039325

z = -0.03727

Using the z score table to find the probability, we have :

P(x<149.75) = 0.48514

In percent = 0.48514 × 100

= 48.514%

Therefore, the probability (as percent) that a random selection of 20 songs will have mean length of less than 149.75 seconds is 48.514%

the answer is 0.071

good luck! : d