The serum cholesterol levels of 12 to 14-year-olds follow a normal distribution with mean 162 mg/dl and standard deviation 28 mg/dl. What percentage of 12 to 14-year-olds have serum cholesterol values
(a) 171 or more?
(b) 143 or less?
(c) 194 or less?
(d) 105 or more?
(e) between 166 and 194?
(f) between 105 and 138?
(g) between 138 and 166?
Solution
The serum cholesterol levels of 12 to 14-year-olds follow a normal distribution with mean 162 mg/dl and standard deviation 28 mg/dl.
Let $X$ denote serum cholesterol levels of 12 to 14-year-olds. Given that $\mu = 162$ and $\sigma = 28$.
(a) The probability that 12 to 14-year-olds have serum cholesterol values 171 or more is
$$ \begin{aligned} P(X\geq171) &= 1-P(X < 171)\\ &= 1- P\bigg(\frac{X-\mu}{\sigma} < \frac{171-162}{28}\bigg)\\ &=1- P\bigg(Z < 0.32\bigg)\\ &=1-0.6261\\ &= 0.3739 \end{aligned} $$
The percentage of 12 to 14-year-olds have serum cholesterol values 171 or more is 37.39.
(b) The probability that 12 to 14-year-olds have serum cholesterol values 143 or less is
$$ \begin{aligned} P(X\leq 143) &= P\bigg(\frac{X-\mu}{\sigma} \leq \frac{143-162}{28}\bigg)\\ &= P\bigg(Z \leq -0.68\bigg)\\ &=0.2487 \end{aligned} $$
The percentage of 12 to 14-year-olds have serum cholesterol values 143 or less is 24.87
(c) The probability that 12 to 14-year-olds have serum cholesterol values 194 or less is
$$ \begin{aligned} P(X\leq 194) &= P\bigg(\frac{X-\mu}{\sigma} \leq \frac{194-162}{28}\bigg)\\ &= P\bigg(Z \leq 1.14\bigg)\\ &=0.8735 \end{aligned} $$
The percentage of 12 to 14-year-olds have serum cholesterol values 194 or less is 87.35
(d) The probability that 12 to 14-year-olds have serum cholesterol values 105 or less is
$$ \begin{aligned} P(X\geq105) &= 1-P(X < 105)\\ &= 1- P\bigg(\frac{X-\mu}{\sigma} < \frac{105-162}{28}\bigg)\\ &=1- P\bigg(Z < -2.04\bigg)\\ &=1-0.0209\\ &= 0.9791 \end{aligned} $$
The percentage of 12 to 14-year-olds have serum cholesterol values 105 or less is 97.91
(e) The probability that 12 to 14-year-olds have serum cholesterol values between 166 and 194 is
$$ \begin{aligned} P(166 < X < 194) &= P\bigg(\frac{166-162}{28}< \frac{X-\mu}{\sigma} < \frac{194-162}{28}\bigg)\\ &=P\bigg(0.143 < Z < 1.143\bigg)\\ &= P(Z < 1.143) -P(Z < 0.143)\\ &=0.8735-0.5568\\ &= 0.3167 \end{aligned} $$
The percentage of 12 to 14-year-olds have serum cholesterol values between 166 and 194 is 31.67
(f) The probability that 12 to 14-year-olds have serum cholesterol values between 105 and 138 is
$$ \begin{aligned} P(105 < X < 138) &= P\bigg(\frac{105-162}{28}< \frac{X-\mu}{\sigma} < \frac{138-162}{28}\bigg)\\ &=P\bigg(-2.036 < Z < -0.857\bigg)\\ &= P(Z < -0.857) -P(Z < -2.036)\\ &=0.1957-0.0209\\ &= 0.1748 \end{aligned} $$
The probability that 12 to 14-year-olds have serum cholesterol values between 105 and 138 is 17.48
(g) The probability that 12 to 14-year-olds have serum cholesterol values between 138 and 166 is
$$ \begin{aligned} P(138 < X < 166) &= P\bigg(\frac{138-162}{28} < \frac{A-\mu}{\sigma} < \frac{166-162}{28}\bigg)\\ &=P\bigg(-0.857 < Z < 0.143\bigg)\\ &= P(Z < 0.143) -P(Z < -0.857)\\ &=0.5568-0.1957\\ &= 0.3611 \end{aligned} $$
The percentage of 12 to 14-year-olds have serum cholesterol values between 138 and 166 is 36.11
Refer to above Question
Suppose a 13-year-old is chosen at random and let Y be the person's serum cholesterol value. Find (a) $P(Y\geq 166)$ (b) $P(166 < Y< 194)$.
(a)
$$ \begin{aligned} P(Y\geq166) &= 1-P(Y < 166)\\ &= 1- P\bigg(\frac{Y-\mu}{\sigma} < \frac{166-162}{28}\bigg)\\ &=1- P\bigg(Z < 0.14\bigg)\\ &=1-0.5568\\ &= 0.4432 \end{aligned} $$
(b)
$$ \begin{aligned} P(166 < Y < 194) &= P\bigg(\frac{166-162}{28} < \frac{Y-\mu}{\sigma} < \frac{194-162}{28}\bigg)\\ &=P\bigg(0.143 < Z < 1.143\bigg)\\ &= P(Z < 1.143) -P(Z < 0.143)\\ &=0.8735-0.5568\\ &= 0.3167 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
