Question

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.

Answer 1

Probability = 0.1039

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

Answer 2

8.53%.

Explanation:

We must calculate the value of z, which is defined by the following formula:

z (x) = (x - m) / [sd / sqrt (n)]

Where x is the value of the variable (188.58), m is the mean (200), sd is the standard deviation (25), and the population size is n (9).

Having the values we replace and we are left with:

z (188.58) = (188.58 - 200) / [25 / sqrt (9)]

z (188.58) = -1.3704

If we look for this value in the table of z (attached), we have that the probability would be: approximately 0.0853 or 8.53%.