Question

Suppose that the TSH (Thyroid Stimulating Hormone) levels among healthy individuals are normally distributed with a mean of 3.3 unitsmL. Suppose also that exactly 98% of healthy individuals have TSH levels below 5.9 unitsmL. Find the standard deviation of the distribution of TSH levels of healthy individuals. Carry your intermediate computations to at least four decimal places. Round your answer to at least two decimal plac

Answer 1

Answer:

The standard deviation of the distribution of TSH levels of healthy individuals is of 1.2658 units/mL.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 3.3 units/mL

This means that $\mu = 3.3$

Suppose also that exactly 98% of healthy individuals have TSH levels below 5.9 units/mL.

This means that when $X = 5.9$, Z has a pvalue of 0.98. So Z when X = 5.9, has a pvalue if 0.98, that is, Z = 2.054. We use this to find $\sigma$

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

$2.054 = \frac{5.9 - 3.3}{\sigma}$

$2.054\sigma = 2.6$

$\sigma = \frac{2.6}{2.054}$

$\sigma = 1.2658$

The standard deviation of the distribution of TSH levels of healthy individuals is of 1.2658 units/mL.