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klasskru [66]
3 years ago
9

Problem PageQuestion Calcium levels in people are normally distributed with a mean of mg/dL and a standard deviation of mg/dL. I

ndividuals with calcium levels in the bottom of the population are considered to have low calcium levels. Find the calcium level that is the borderline between low calcium levels and those not considered low. Carry your intermediate computations to at least four decimal places. Round your answer to one decimal place.

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
6 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  calcium level that is the borderline between low calcium levels and those not considered low is  c = 8.68

Step-by-step explanation:

From the question we are told that

    The mean is  \mu =  9.5\  mg/dL

     The standard deviation \sigma =  0.5 \ mg/dL

    The proportion of the population with low calcium level is  p =5% = 0.05

Let X be a X random calcium level

  Now the  P(X < c) = 0.05

Here P denotes probability

        c is population with calcium level at the borderline

  Since the calcium level is normally distributed the z-value is  evaluated as

    P(Z < \frac{c - \mu}{\sigma } ) = 0.05

The critical value for 0.05 from the standard normal distribution table is

       t_{0.05} = -1.645

=>    \frac{c - \mu}{\sigma }  =  -1.645

substituting values

       \frac{c - 9.5}{0.5 }  =  -1.645

=>    c - 9.5 =  -0.8225

=>     c = 8.68

 

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Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

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w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

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----------------------

Based on that root, we need to try a sample value that is to the left of it.

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Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

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