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).

Question

What is the CORRECT match between the air pollutants and treatment techniques given in the table

Air Pollutants		Treatment techniques
P	NO₂	i	Flaring
Q	SO₂	ii	Cyclonic separator
R	CO	iii	Lime scrubbing
S	Particles	iv	NH₃ injection

Answer 1

Both age and height follow a normal (bell curve) distribution, just with different centers and spreads. The question is really asking: if we go the SAME number of standard deviations above the mean of age as we go above the mean of height, do we land on values with the same probability of being exceeded? Yes, because the shape of every normal curve, once you standardize it, is identical, so equal probabilities always correspond to equal "number of standard deviations from the mean".

How far is 50 years above the mean age, in standard deviations? The mean age is 36 and one standard deviation is 12 years. The distance from the mean to 50 is $50 - 36 = 14$ years. Dividing by the standard deviation, $14/12 \approx 1.1667$ standard deviations above the mean.
What does "equal probability" require for height? Since $P(\text{age} > 50)$ must equal $P(\text{height} > h)$, and both age and height are normal, the point $h$ has to sit the SAME number of standard deviations, $1.1667$, above the mean height.
Convert that many standard deviations of height back into cm. One standard deviation of height is 10 cm, so $1.1667$ standard deviations is $1.1667 \times 10 \approx 11.67$ cm above the mean height of 160 cm.

Adding that to the mean height: $h = 160 + 11.67 = 171.67$ cm (rounded to two decimal places).

Let's summarize:

50 years is about $1.1667$ standard deviations above the mean age.
Matching that same distance in standard deviations for height, above a mean of 160 cm with a spread of 10 cm, gives $h \approx 171.67$ cm.

So the required height is about 171.67 cm.

Answer 2

Both age and height follow a normal (bell curve) distribution, just with different centers and spreads. The question is really asking: if we go the SAME number of standard deviations above the mean of age as we go above the mean of height, do we land on values with the same probability of being exceeded? Yes, because the shape of every normal curve, once you standardize it, is identical, so equal probabilities always correspond to equal "number of standard deviations from the mean".

How far is 50 years above the mean age, in standard deviations? The mean age is 36 and one standard deviation is 12 years. The distance from the mean to 50 is $50 - 36 = 14$ years. Dividing by the standard deviation, $14/12 \approx 1.1667$ standard deviations above the mean.
What does "equal probability" require for height? Since $P(\text{age} > 50)$ must equal $P(\text{height} > h)$, and both age and height are normal, the point $h$ has to sit the SAME number of standard deviations, $1.1667$, above the mean height.
Convert that many standard deviations of height back into cm. One standard deviation of height is 10 cm, so $1.1667$ standard deviations is $1.1667 \times 10 \approx 11.67$ cm above the mean height of 160 cm.

Adding that to the mean height: $h = 160 + 11.67 = 171.67$ cm (rounded to two decimal places).

Let's summarize:

50 years is about $1.1667$ standard deviations above the mean age.
Matching that same distance in standard deviations for height, above a mean of 160 cm with a spread of 10 cm, gives $h \approx 171.67$ cm.

So the required height is about 171.67 cm.

Show Hint

Correct Answer: 171.67

Solution and Explanation

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