Respuesta :
Answer:
0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.
Step-by-step explanation:
The following information is missing:
The standard deviation of population is 100.
We are given the following information in the question:
Population mean, μ = 502
Standard Deviation, σ = 100
Sample size, n = 90
Standard error =
[tex]\dfrac{\sigma}{\sqrt{n}} = \dfrac{100}{\sqrt{90}} = 10.54[/tex]
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(test score within 10 points)
[tex]P(492 \leq x \leq 512) \\\\= P(\displaystyle\frac{492 - 502}{10.54} \leq z \leq \displaystyle\frac{512-502}{10.54}) \\\\= P(-0.9487 \leq z \leq 0.9487)\\= P(z \leq 0.9487) - P(z < -0.9487)\\= 0.829 -0.171 = 0.658 = 65.8\%[/tex]
[tex]P(492 \leq x \leq 512) = 65.8\%[/tex]
0.658 is the probability that a sample 90 test takers will provide a sample mean test score within 10 points of the population mean of 502.
