Answer:
0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8 minutes
Standard Deviation, σ = 2.5 minutes
Sample size, n = 25
We are given that the distribution of time spent is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling =
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{2.5}{\sqrt{25}} = 0.5[/tex]
P(sample mean is between 7.8 and 8.2 minutes)
[tex]P(7.8 \leq x \leq 8.2)\\\\ = P(\displaystyle\frac{7.8 - 8}{0.5} \leq z \leq \displaystyle\frac{8.2-8}{0.5})\\\\ = P(-0.4 \leq z \leq 0.4})\\\\= P(z < 0.4) - P(z < -0.4)\\\\= 0.6554 -0.3446= 0.3108[/tex]
0.3108 is the probability that the sample mean is between 7.8 and 8.2 minutes.
