A psychologist has devised a stress test for dental patients sitting in the waiting rooms. According to this test, the stress scores (on a scale of 1 to 10) for patients waiting for root canal treatments are found to be approximately normally distributed with a mean of 7.59 and a standard deviation of .73.

a. What percentage of such patients have a stress score lower than 6.0?
b. What is the probability that a randomly selected root canal patient sitting in the waiting room has a stress score between 7.0 and 8.0?
c. The psychologist suggests that any patient with a stress score of 9.0 or higher should be given a sedative prior to treatment. What percentage of patients waiting for root canal treatments would need a sedative if this suggestion is accepted?

Solution

Given that $\mu = 7.59$ and $\sigma = 0.73$. $X\sim N(7.59, 0.73^2)$.

a. Probability that a patients have a stress score lower than 6.0 is

$$ \begin{aligned} P(X < 6) &= P\bigg(\frac{X-\mu}{\sigma} < \frac{6-7.59}{0.73}\bigg)\\ &= P(Z < -2.178)\\ &=0.0147 \end{aligned} $$

1.47 percent of such patients have a stress score lower than 6.0.

Normal Distribution
Normal Distribution

b. The probability that a randomly selected root canal patient sitting in the waiting room has a stress score between 7.0 and 8.0 is

$$ \begin{aligned} P(7 < X< 8)&= P\bigg(\frac{7-7.59}{0.73}< \frac{X-\mu}{\sigma} < \frac{8-7.59}{0.73}\bigg)\\ &=P\bigg(-0.808 < Z < 0.562\bigg)\\ &= P(Z < 0.562) -P(Z < -0.808)\\ &=0.7129-0.2095\\ &= 0.5034 \end{aligned} $$

Normal Distribution
Normal Distribution

c. The probability that patients with stress score of 9.0 or higher is

$$ \begin{aligned} P(X > 9) &= P(X > 9)\\ &=P\bigg(\frac{X-\mu}{\sigma} > \frac{9-7.59}{0.73}\bigg)\\ &=P\bigg(Z > 1.932\bigg)\\ &= 1- P(Z\leq 1.932)\\ &=1-0.9733\\ &= 0.0267 \end{aligned} $$

Normal Distribution
Normal Distribution

If the suggestion is accepted, then $2.67$ percent of patients waiting for root canal treatments would need a sedative.