If the length of country and western songs has mean 170 seconds and standard deviation 30 seconds, then the probability that a random selection of 16 songs will have mean length of 160.48 seconds or less is
A) 0.8980
B) 0.3980
C) 0.1020
D) 0.6020

Question

Quizplus · Accepted Answer

We are given:

$\mu = 170$ seconds
$\sigma = 30$ seconds
$n = 16$ songs
$\bar{x} = 160.48$ seconds

We are asked to find the probability that a random selection of 16 songs will have a mean length of 160.48 seconds or less. We can solve this by using the central limit theorem.

The central limit theorem states that if we have a sample size $n$ that is large enough (usually $n \geq 30$), then the distribution of the sample means will be approximately normal.

The mean of the sample means ($\mu_{\bar{x}}$) is the same as the population mean ($\mu$), which is 170 seconds. The standard deviation of the sample means ($\sigma_{\bar{x}}$) is equal to the population standard deviation divided by the square root of the sample size:

$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{30}{\sqrt{16}} = 7.5$ seconds

We can now standardize the sample mean using the z-score formula:

$z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{160.48 - 170}{7.5} = -1.27$

Using a standard normal distribution table, we can find the probability that a randomly selected sample mean will be less than or equal to $\bar{x} = 160.48$:

$P(\bar{x} \leq 160.48) = P(z \leq -1.27) = 0.102$

Therefore, the probability that a random selection of 16 songs will have a mean length of 160.48 seconds or less is 0.102 or approximately 10.20%. The correct answer is C.

If the Length of Country and Western Songs Has Mean