Question

Please help me with this problem. I will give brainliest!

Answer 1

Answer:

The number of people in one session that will spend within two standard deviations below the mean and one standard deviations above the mean time on Facespace is 394 people

Step-by-step explanation:

The given information are;

The mean time spent of Facespace, μ = 30 minutes

The standard deviation of the time spent daily, σ = 6 minutes

The number of people in one sitting, n = 2900 people

The time spent two standard deviations below the mean = 30 - 12 = 18 minutes

The time spent one standard deviations above the mean = 30 + 6 = 36 minutes

The Z-score values are;

$Z=\dfrac{x-\mu }{\sigma }$

Which gives;

For x = 30

$Z=\dfrac{36-30 }{6 } = 1$

For x = 18

$Z=\dfrac{18-30 }{6 } = -2$

From the z-score table, we have;

P(Z > -2) = 1 - 0.02275 = 0.97725

P(Z < 1) = 0.84134

Therefore, the probability P(-2 < Z < 1) = 0.97725  - 0.84134 = 0.13591

Given that there are 2900 are on in one sitting, the number of them that will lie within two standard deviations below the mean and one standard deviations above the mean = 2900 × 0.13591 = 394.139 which is approximately 394 people.