# How do you solve this??? need help?

The length of country and western songs is normally distributed and has a mean of 190 seconds and a standard deviation of 30 seconds. find the probability that a random selection of 16 songs will have mean length of 181.98 seconds or less. Assume the distribution of the lengths of the songs is normal.

### 2 Answers

- Ian HLv 72 months ago
With a sample of size n, the role of standard deviation σ is replaced by

the standard error of the mean, σ/√(n) = 30/√(16) = 7.5 in this example.

181.98 is 8.02 secs to the left of the mean 190. In terms of this new type of SD

that is -8.02/7.5 ~ -1.069 “standard deviations”

We are asked about the left tail probability that our random selection of 16 songs will have mean length of 181.98 seconds or LESS.

Using this interactive calculator, (set to UP TO)

https://www.mathsisfun.com/data/standard-normal-di...

P(z < -1.07) ~ 14.23%

- Wayne DeguManLv 72 months ago
We have P(X < x) => P[Z < (X - µ)/σ]

With a sample of size n, this becomes:

P[Z < (X - µ)/(σ/√n)]

so, P[Z < (181.98 - 190)/(30/√16)]

i.e. P[Z < -8.02/7.5]

=> P(z < -1.07)

or, P(z > 1.07)...by symmetry

so, 1 - P(z < 1.07)

Using normal tables we get:

1 - 0.8577 = 0.1423

:)>