Question

Arsenic is a compound that occurs naturally in very low concentrations. Arsenic blood concentrations in healthy adults are Normally distributed with mean =3.2 micrograms per deciliter (ug/dl) and standard deviation sigma = 1.5. What is the range of arsenic blood concentrations corresponding to the middle 90% of healthy adults?

Answer 1

Answer:

0.7325 to 5.6675 ug/dl

Step-by-step explanation:

The middle 90% will be 45% above the mean and 45% below the mean.  This means

0.5-0.45 = 0.05 and

0.5+0.45 = 0.95

We use a z table.  Look in the cells; find the values as close to 0.05 and 0.95 as we can get.

For 0.05, we have 0.0505 and 0.0495; since these are equidistant from 0.05, we use the value between them.  0.0505 is z=-1.64 and 0.0495 is z=1.65; this gives us z=-1.645.

For 0.95, we have 0.9495 and 0.9505; since these are equidistant from 0.95, we use the value between them.  0.9495 is z = 1.64 and 0.9505 is z=1.65; this gives us z = 1.645.

Now we use our z score formula,

$z=\frac{X-\mu}{\sigma}$

Our two z scores are 1.645 and -1.645; our mean, μ, is 3.2; and our standard deviation, σ, is 1.5:

$1.645=\frac{X-3.2}{1.5}$

Multiply both sides by 1.5:

$1.5(1.645)=\frac{X-3.2}{1.5}\times 1.5\\\\2.4675 = X-3.2$

Add 3.2 to each side:

2.4675+3.2 = X-3.2+3.2

5.6675 = X

$-1.645=\frac{X-3.2}{1.5}$

Multiply both sides by 1.5:

$1.5(-1.645)=\frac{X-3.2}{1.5}\times 1.5\\\\-2.4675=X-3.2$

Add 3.2 to each side:

-2.4675+3.2 = X-3.2+3.2

0.7325 = X

Our range is from 0.7325 to 5.6675.

Answer 2

Using the normal distribution, it is found that the middle 90% is between 0.26 and 6.14 ug/dl.

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• In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is of 3.5 ug/dl, thus $\mu = 3.5$.
• The standard deviation is of 1.5 ug/dl, thus $\sigma = 1.5$.
• The normal distribution is symmetric, which means that the middle 90% is between the 5th percentile and the 95th percentile.
• The 5th percentile is X when Z has a p-value of 0.05, so X when Z = -1.96.
• The 95th percentile is X when Z has a p-value of 0.95, so X when Z = 1.96.

5th percentile:

$Z = \frac{X - \mu}{\sigma}$

$-1.96 = \frac{X - 3.2}{1.5}$

$X - 3.2 = -1.96(1.5)$

$X = 0.26$

95th percentile:

$Z = \frac{X - \mu}{\sigma}$

$1.96 = \frac{X - 3.2}{1.5}$

$X - 3.2 = 1.96(1.5)$

$X = 6.14$

The range of the middle 90% is between 0.26 and 6.14 ug/dl.

A similar problem is given at https://brainly.com/question/23285700