The length of country and western songs is normally distributed and has a mean of 200seconds and a standard deviation of 25 seconds. Find the probability that a randomselection of 9 songs will have mean length of 188.58 seconds or less. Assume thedistribution of the lengths of the songs is normal.

Probability = 0.1039

Step-by-step explanation:

To solve this problem, we have to find the t - value.

From the question,

Mean; μ = 200

Standard deviation; σ = 25

N = 9

Distribution of the t statistic (also known as the t score), whose values are given by:

t (x) = (x - μ) / [σ/√ (n) ]

x = sample mean

μ = Population mean

n = sample size

σ = standard deviation

In this question,

x = 188.58

μ = 200

n = 9

σ = 25

Plugging in these values, we now have;

t (188.58) = (188.58 - 200) / [25/√ (9) ]

t (188.58) = - 1.3704

Now, Degree of Freedom is given by,

DF = n - 1

So, DF = 9 - 1 = 8

So, for P (x < 188.58) with degree of freedom of 8 and and t value as - 1.3704, from the t-distribution table and online calculator, we have a value of 0.1039