Question 1168947: On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas, considered safe. Suppose that the standard deviation is 1.4 hours and the amount of time spent alone is normally distributed. We randomly survey one Mercurian 4-year-old living in a rural area. We are interested in the amount of time X the child spends alone per day. (Source: San Jose Mercury News) Round all answers to 4 decimal places where possible.
a. What is the distribution of X?
b. Find the probability that the child spends less than 2.8 hours per day unsupervised.
c. What percent of the children spend over 3.9 hours per day unsupervised. (Round to 2 decimal places)
d. 78% of all children spend at least how many hours per day unsupervised?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the amount of time a randomly surveyed Mercurian 4-year-old living in a rural area spends alone per day. We are given that $X$ is normally distributed with a mean $\mu = 3$ hours and a standard deviation $\sigma = 1.4$ hours.
**a. What is the distribution of X?**
Since the problem states that the amount of time spent alone is normally distributed with a mean of 3 hours and a standard deviation of 1.4 hours, the distribution of $X$ is:
$X \sim N(3, 1.4^2)$ or $X \sim N(3, 1.96)$
In terms of mean and standard deviation, we write:
$X \sim N(\mu=3, \sigma=1.4)$
**b. Find the probability that the child spends less than 2.8 hours per day unsupervised.**
We want to find $P(X < 2.8)$. To do this, we first calculate the z-score:
$z = \frac{x - \mu}{\sigma} = \frac{2.8 - 3}{1.4} = \frac{-0.2}{1.4} \approx -0.142857$
Now we find the probability $P(Z < -0.1429)$ using a standard normal distribution table or a calculator.
$P(Z < -0.1429) \approx 0.4432$
The probability that the child spends less than 2.8 hours per day unsupervised is approximately 0.4432.
**c. What percent of the children spend over 3.9 hours per day unsupervised. (Round to 2 decimal places)**
We want to find $P(X > 3.9)$. First, we calculate the z-score:
$z = \frac{x - \mu}{\sigma} = \frac{3.9 - 3}{1.4} = \frac{0.9}{1.4} \approx 0.642857$
Now we find the probability $P(Z > 0.6429) = 1 - P(Z < 0.6429)$ using a standard normal distribution table or a calculator.
$P(Z < 0.6429) \approx 0.7399$
$P(Z > 0.6429) = 1 - 0.7399 = 0.2601$
To find the percentage, we multiply the probability by 100:
$0.2601 \times 100 = 26.01\%$
Approximately 26.01% of the children spend over 3.9 hours per day unsupervised. Rounded to 2 decimal places, this is 26.01%.
**d. 78% of all children spend at least how many hours per day unsupervised?**
We are looking for the value $x$ such that $P(X \ge x) = 0.78$. This is equivalent to finding the value $x$ such that $P(X < x) = 1 - 0.78 = 0.22$.
First, we find the z-score corresponding to a cumulative probability of 0.22. Using a standard normal distribution table or a calculator, we find the z-score $z$ such that $P(Z < z) = 0.22$.
$z \approx -0.7722$
Now we use the z-score formula to find the corresponding value of $x$:
$z = \frac{x - \mu}{\sigma}$
$-0.7722 = \frac{x - 3}{1.4}$
$x - 3 = -0.7722 \times 1.4$
$x - 3 = -1.08108$
$x = 3 - 1.08108$
$x = 1.91892$
Rounding to 4 decimal places, $x \approx 1.9189$ hours.
78% of all children spend at least approximately 1.9189 hours per day unsupervised.
Final Answer: The final answer is $\boxed{X \sim N(3, 1.4), 0.4432, 26.01\%, 1.9189}$
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