Identify ALL solutions to the inequality. There could be more than one selection.
2 answers:
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Answer:
The probability of healthcare expenses in the population being greater than $4,000 is 0.02275.
Step-by-step explanation:
We are given that yearly healthcare expenses for a family of four are normally distributed with a mean expense equal to $3,000 and a standard deviation equal to $500.
Let X = <u><em>yearly healthcare expenses of a family</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean expense = $3,000
= standard deviation = $500
Now, the probability of healthcare expenses in the population being greater than $4,000 is given by = P(X > $4,000)
P(X > $4,000) = P( >
) = P(Z > 2) = 1 - P(Z
2)
= 1 - 0.97725 = <u>0.02275</u>
The above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.
The second question is (x-5)(2x+7)
