Answer:
The answer is below
Step-by-step explanation:
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by the formula:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mean,\ \sigma=standard\ deviation.\\\\For \ a\ sample\ size(n):\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]
Therefore given that μ = $85900, σ = $11000, n = 35.
For x > 70000:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } =\frac{70000-85900}{11000/\sqrt{35} } =-8.55\\[/tex]
For x < 100000
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } =\frac{100000-85900}{11000/\sqrt{35} } =7.58[/tex]
From the normal distribution table, P(70000< x < 100000) = P(-8.55 < z < 7.58) = P(z < 7.58) - P(z < -8.55) = 1 - 0 = 1 = 100%
