Let X be normally distributed with mean μ = 2,800 and standard deviation σ = 900[You may find it useful to reference the z table.]
a. Find x such that P(X ≤ x) = 0.9382. (Round "z" value to 2 decimal places, and final answer to nearest whole number.)
b. Find x such that
P(X > x) = 0.025. (Round
"z" value to 2 decimal places, and final answer to nearest
whole number.)
c. Find x such that P(2,800 ≤
X ≤ x) = 0.1217. (Round
"z" value to 2 decimal places, and final answer to
nearest whole number.)
d. Find x such that P(X
≤ x) = 0.4840. (Round
"z" value to 2 decimal places,
and final answer to nearest whole number.)
Get Answers For Free
Most questions answered within 1 hours.