In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5. (a) Two adults who both work are chosen at random. Find the probability that at least one of them earns between $15 and $30 per hour. (b) The number of adults who work is equal to three times the number of adults who do not work. One adult is chosen at random. Find the probability that the chosen adult works and earns more than $25 per hour.

SSCE HSC Mathematics Advanced Absolute value functions: In a particular country, the hou

In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5 - HSC - SSCE Mathematics Advanced - Question 28 - 2020 - Paper 1

Question 28

In-a-particular-country,-the-hourly-rate-of-pay-for-adults-who-work-is-normally-distributed-with-a-mean-of-$25-and-a-standard-deviation-of-$5-HSC-SSCE Mathematics Advanced-Question 28-2020-Paper 1.png

In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5. (a) Two adults who bo... show full transcript

Worked Solution & Example Answer:In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5 - HSC - SSCE Mathematics Advanced - Question 28 - 2020 - Paper 1

Step 1

Find the probability that at least one earns between $15 and $30 per hour.

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Answer

To find the probability that at least one of the two adults earns between $15 and$ 30, we start by determining the probabilities for individuals. The given mean is $25, and the standard deviation is$ 5.

Calculate the Z-scores:
- For $15: Z = \frac{15 - 25}{5} = -2$ .
- For $30: Z = \frac{30 - 25}{5} = 1$ .
Find cumulative probabilities:
- From standard normal distribution tables, we have:
  - $P(Z < -2) \approx 0.0228$ .
  - $P(Z < 1) \approx 0.8413$ .
Find the probability of earning between $15 and$ 30:
- $P(15 < X < 30) = P(Z < 1) - P(Z < -2) = 0.8413 - 0.0228 = 0.8185$ .
Find probability of neither earning between $15 and$ 30:
- $P(neither earns between$ 15 and $30) = (1 - 0.8185)^2 = (0.1815)^2 = 0.0329$ .
Use complement rule:
- $P(at least one earns between$ 15 and $30) = 1 - P(neither earns between$ 15 and $30) = 1 - 0.0329 \approx 0.9671$ .

Step 2

Find the probability that the chosen adult works and earns more than $25 per hour.

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Answer

Given that the number of adults who work is equal to three times the number that do not work, we have:

Let the number of adults who do not work be $x$ .
Then, the number of adults who work is $3x$ .
Total number of adults = $4x$ .

The probability that a randomly chosen adult works is: $P(\text{chosen adult works}) = \frac{3x}{4x} = \frac{3}{4}$ .

Next, we need to find the probability that an adult earns more than $25 per hour:

Since earnings are normally distributed with a mean of $25, the probability that an adult earns more than$ 25 is: $P(X > 25) = 1 - P(X \leq 25) = 0.5$ (since $P(X \leq 25) = 0.5$ for a normal distribution).

Finally, the combined probability that the chosen adult works and earns more than $25 per hour is: $P(\text{adult works and earns more than } 25) = P(\text{chosen adult works}) \times P(X > 25) = \frac{3}{4} \times 0.5 = \frac{3}{8}.$

In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5 - HSC - SSCE Mathematics Advanced - Question 28 - 2020 - Paper 1

Worked Solution & Example Answer:In a particular country, the hourly rate of pay for adults who work is normally distributed with a mean of $25 and a standard deviation of $5 - HSC - SSCE Mathematics Advanced - Question 28 - 2020 - Paper 1

Find the probability that at least one earns between $15 and $30 per hour.

Find the probability that the chosen adult works and earns more than $25 per hour.

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