Question

The age (in years) of a population is normally distributed with a mean of 36 and standard deviation of 12. The height (in cm) of the same population is also normally distributed with a mean of 160 and standard deviation of 10.
If the probability of age greater than 50 years is equal to the probability of height greater than \(h\), the value of \(h\) (in cm) is ______ (rounded off to two decimal places).

Hint:

Equal tail probabilities in two normal distributions mean equal Z-scores: \(\dfrac{50-36}{12}=\dfrac{h-160}{10}\).

Answer 1

Correct Answer: 171.67

Step 1: Set up the standard normal (Z) variable for age.
Age is normally distributed with mean \(\mu_1=36\) years and standard deviation \(\sigma_1=12\) years.
For any normal variable, we convert a raw value to a Z-score using \(Z=\dfrac{x-\mu}{\sigma}\), which tells us how many standard deviations that value is away from the mean. This lets us compare two DIFFERENT normal distributions (age and height) on the same scale.

Step 2: Find the Z-score for "age greater than 50".
\[ Z_{age}=\frac{50-36}{12}=\frac{14}{12}\approx1.1667 \] So \(P(\text{age}>50)=P(Z>1.1667)\), using the standard normal distribution.

Step 3: Set up the Z-score for height in terms of h.
Height is normally distributed with mean \(\mu_2=160\) cm and standard deviation \(\sigma_2=10\) cm.
\[ Z_{height}=\frac{h-160}{10} \] So \(P(\text{height}>h)=P\!\left(Z>\dfrac{h-160}{10}\right)\).

Step 4: Equate the two probabilities.
Since both events are of the form "a normal variable exceeds a value", and we are told these two probabilities are equal, the two Z-scores that mark those tail areas must themselves be equal (the same tail area under the standard normal curve corresponds to a unique Z value).
\[ \frac{h-160}{10}=\frac{14}{12} \]
Step 5: Solve for h.
\[ h-160=10\times\frac{14}{12}=\frac{140}{12}\approx11.67 \] \[ h=160+11.67=171.67 \text{ cm (rounded to two decimal places)} \]
Final Answer:
The value of h is about 171.67 cm. \[ \boxed{h\approx171.67 \text{ cm}} \]