A recent survey asked 1682 people how many hours per day they were able to relax. The results are presented in the following table.

Question

Number of Hours: 0, 1, 2, 3, 4, 5, 6, 7,8
frequency: 113, 184, 336, 251, 318, 234, 151, 35, 60 for a total of 1682
Consider these 1682 people to be a population. Let X be the number of hours of relaxation for a person sampled at random from this population.
(a) Construct the probability distribution of X. Round the answer to three decimal places.
(b) Find the probability that a person relaxes more than 5 hours per day.
(c) Find the probability that a person doesn't relax at all.
(d) Compute the mean UvX. Round the answer to two decimal places.
(e) Compute the standard deviation OvX. Round the answer to three decimal places.

Answer 1

To address the problem, let's first calculate the probability distribution based on the given frequencies.

(a) Construct the probability distribution of X.

The probability $P(X=x)$ can be calculated using the formula:

$$P(X=x)=\frac{\text{Frequency of }x}{\text{Total Frequency}}=\frac{\text{Frequency of }x}{1682}$$

Let's calculate the probabilities for each number of hours:

| Number of Hours (x) | Frequency | P(X = x) | |----------------------|-----------|------------------------| | 0 | 113 | $\frac{113}{1682}=0.067$ | | 1 | 184 | $\frac{184}{1682}=0.109$ | | 2 | 336 | $\frac{336}{1682}=0.200$ | | 3 | 251 | $\frac{251}{1682}=0.149$ | | 4 | 318 | $\frac{318}{1682}=0.189$ | | 5 | 234 | $\frac{234}{1682}=0.139$ | | 6 | 151 | $\frac{151}{1682}=0.090$ | | 7 | 35 | $\frac{35}{1682}=0.021$ | | 8 | 60 | $\frac{60}{1682}=0.036$ |

Thus, the probability distribution of $X$ is:

$$\begin{array}{rlrlrlrlrl}P(X=0)& =0.067P(X=1)& =0.109P(X=2)& =0.200P(X=3)& =0.149P(X=4)& =0.189P(X=5)& =0.139P(X=6)& =0.090P(X=7)& =0.021P(X=8)& =0.036\end{array}$$

(b) Find the probability that a person relaxes more than 5 hours per day.

To find $P(X>5)$:

$$P(X>5)=P(X=6)+P(X=7)+P(X=8)$$

Calculating this:

$$P(X>5)=0.090+0.021+0.036=0.147$$

(c) Find the probability that a person doesn't relax at all.

This is simply $P(X=0)$:

$$P(X=0)=0.067$$

(d) Compute the mean ${\mu }_X$.

The mean ${\mu }_X$ is calculated using:

$${\mu }_X=\sum (x\cdot P(X=x))$$

Let's compute this:

$$\begin{array}{rlrl}{\mu }_X& =0\cdot 0.067+1\cdot 0.109+2\cdot 0.200+3\cdot 0.149+4\cdot 0.189+5\cdot 0.139+6\cdot 0.090+7\cdot 0.021+8\cdot 0.036& =0+0.109+0.400+0.447+0.756+0.695+0.540+0.147+0.288& =3.382\end{array}$$

Thus, the mean ${\mu }_X$ rounded to two decimal places is:

$${\mu }_X\approx 3.38$$

(e) Compute the standard deviation ${\sigma }_X$.

First, calculate the variance ${\sigma }_X^2$:

$${\sigma }_X^2=\sum (x^2\cdot P(X=x))-{\mu }_X^2$$

To calculate this, we need $x^2\cdot P(X=x)$:

$$\begin{array}{rlrlrlrlrl}{0}^2\cdot 0.067& =01^2\cdot 0.109& =0.1092^2\cdot 0.200& =0.8003^2\cdot 0.149& =1.3414^2\cdot 0.189& =3.0245^2\cdot 0.139& =1.7356^2\cdot 0.090& =3.2407^2\cdot 0.021& =1.0298^2\cdot 0.036& =2.304\end{array}$$

Now sum these products:

$$0+0.109+0.800+1.341+3.024+1.735+3.240+1.029+2.304=13.582$$

Now, plug this into the variance formula:

$${\sigma }_X^2=13.582-(3.382{)}^2=13.582-11.438=2.144$$

Finally, calculate the standard deviation:

$${\sigma }_X=\sqrt{2.144}\approx 1.464$$

Thus, the standard deviation ${\sigma }_X$ rounded to three decimal places is:

$${\sigma }_X\approx 1.464$$

Summary of Answers:

(a) Probability distribution: given above.
(b) $P(X>5)=0.147$
(c) $P(X=0)=0.067$
(d) Mean ${\mu }_X\approx 3.38$
(e) Standard deviation ${\sigma }_X\approx 1.464$

Answer 2

E is wrong

Answer 3

Let's re-calculate the standard deviation step by step, making sure we address any inaccuracies in the previous calculations.

Given Data Recap:

The probability distribution was calculated correctly:

| Number of Hours (x) | Frequency | Probability $P(X=x)$ | |----------------------|-----------|-----------------------------| | 0 | 113 | 0.067 | | 1 | 184 | 0.109 | | 2 | 336 | 0.200 | | 3 | 251 | 0.149 | | 4 | 318 | 0.189 | | 5 | 234 | 0.139 | | 6 | 151 | 0.090 | | 7 | 35 | 0.021 | | 8 | 60 | 0.036 |

(d) Recall Mean Calculation

To recalculate the mean ${\mu }_X$:

$${\mu }_X=\sum (x\cdot P(X=x))$$

Calculating:

$$\begin{array}{rlrl}{\mu }_X& =0\cdot 0.067+1\cdot 0.109+2\cdot 0.200+3\cdot 0.149+4\cdot 0.189+5\cdot 0.139+6\cdot 0.090+7\cdot 0.021+8\cdot 0.036& =0+0.109+0.400+0.447+0.756+0.695+0.540+0.147+0.288& =3.382\end{array}$$

The mean ${\mu }_X\approx 3.38$, which appears correct.

(e) Recalculate Standard Deviation ${\sigma }_X$

1. Calculate $\sum (x^2\cdot P(X=x))$:

$$\begin{array}{rlrlrlrlrl}{0}^2\cdot P(X=0)& =0\cdot 0.067=01^2\cdot P(X=1)& =1\cdot 0.109=0.1092^2\cdot P(X=2)& =4\cdot 0.200=0.8003^2\cdot P(X=3)& =9\cdot 0.149=1.3414^2\cdot P(X=4)& =16\cdot 0.189=3.0245^2\cdot P(X=5)& =25\cdot 0.139=3.4756^2\cdot P(X=6)& =36\cdot 0.090=3.2407^2\cdot P(X=7)& =49\cdot 0.021=1.0298^2\cdot P(X=8)& =64\cdot 0.036=2.304\end{array}$$

Now sum these contributions: $$0+0.109+0.800+1.341+3.024+3.475+3.240+1.029+2.304=15.322$$

2. Compute variance: $${\sigma }_X^2=\sum (x^2\cdot P(X=x))-{\mu }_X^2=15.322-(3.382{)}^2$$ Calculating $(3.382{)}^2$: $$(3.382{)}^2\approx 11.438$$

Thus: $${\sigma }_X^2=15.322-11.438\approx 3.884$$

3. Standard Deviation: $${\sigma }_X=\sqrt{3.884}\approx 1.970$$

Therefore, the standard deviation ${\sigma }_X$ rounded to three decimal places is:

$${\sigma }_X\approx 1.970$$

Summary of Corrected Answers:

• (d) Mean ${\mu }_X\approx 3.38$ (unchanged)
• (e) Standard deviation ${\sigma }_X\approx 1.970$ (corrected)

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