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FromTheMoon [43]

2 years ago

11

The Marsh family took a vacation that covered a total distance of 1356 miles. The return trip was 284 miles shorter than the fir

st part of the trip. How long was the return trip?

1 answer:

Alex787 [66]2 years ago

6 0

So going+return=1356
return is 284 lesss than going
return=-284+going
subsitute

going-284+going=1356
2going-284=1356
add 284 to both sides
2going=1640
divide both sides by 2
going=820

so we havve
return=-284+going
return=-284+820
return=536

answer is return=536 miles

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Each of 16 students measured the circumference of a tennis ball by four different methods, which were:Method A: Estimate the cir

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Answer:

a) $\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

b) $Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

c) $\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

Step-by-step explanation:

Data given

Assuming the following data:

Method A: 18.0, 18.0, 18.0, 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0,25.0,26.0,26.4

Method B: 18.8, 18.9, 18.9, 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0, 22.0,22.0,23.6

Method C: 20.2, 20.5, 20.5, 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5,21.5,21.5,21.6

Method D: 20.0, 20.0, 20.0,-20.0, 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5, 21.6, 22.1,22.3.

Part a

The sample mean is defined as:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

If we apply this formula for the four methods we got:

$\bar X_A =22.744$

$\bar X_B =20.7$

$\bar X_C =21.013$

$\bar X_D =18.306$

Part b

Since the total number of points for each method is 16 and this is an even number we need to calculate the median as the average between the 8th and the 9th position of the data ordered from the smallest to the largest. If we do this we have that:

$Median_A =\frac{23+24}{2}=23.5$

$Median_B =\frac{20.4+20.4}{2}=20.4$

$Median_C =\frac{21+21}{2}=21$

$Median_D =\frac{20.7+20.7}{2}=20.7$

Part c

The trimmed mean by 20% means that we need to calculate removing the 20% from each of the tails, the 20% of 16 is 3.2, so we need to remove 3 observations from both ends of the data like this:

Method A: 20.0, 22.0, 22.0, 22.5, 23.0, 24.0,24.0,25.0,25.0, 25.0

Method B: 19.6, 20.1, 20.4, 20.4, 20.4, 20.4,20.5, 21.2. 22.0,22.0

Method C: 20.7, 20.8, 20.9, 21.0, 21.0,21.0,21.0, 21.0, 21.5,21.5

Method D: 20.0 20.2, 20.5, 20.5, 20.7, 20.7,20.7, 21.0, 21.1, 21.5.

If we calculate the mean for each method we got:

$\bar X_A =23.25$

$\bar X_B =20.7$

$\bar X_C =21.04$

$\bar X_D =20.69$

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2 years ago

g Assume that the distribution of time spent on leisure activities by adults living in household with no young children is norma

OLga [1]

Answer:

"<em>The probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day</em>" is about 0.8749.

Step-by-step explanation:

We have here a <em>random variable</em> that is <em>normally distributed</em>, namely, <em>the</em> <em>time spent on leisure activities by adults living in a household with no young children</em>.

The normal distribution is determined by <em>two parameters</em>: <em>the population mean,</em> $\\ \mu$ , and <em>the population standard deviation,</em> $\\ \sigma$ . In this case, the variable follows a normal distribution with parameters $\\ \mu = 4.5$ hours per day and $\\ \sigma = 1.3$ hours per day.

We can solve this question following the next strategy:

Use the <em>cumulative</em> <em>standard normal distribution</em> to find the probability.
Find the <em>z-score</em> for the <em>raw score</em> given in the question, that is, <em>x</em> = 6 hours per day.
With the <em>z-score </em>at hand, we can find this probability using a table with the values for the <em>cumulative standard normal distribution</em>. This table is called the <em>standard normal table</em>, and it is available on the Internet or in any Statistics books. Of course, we can also find these probabilities using statistics software or spreadsheets.

We use the <em>standard normal distribution </em>because we can "transform" any raw score into <em>standardized values</em>, which represent distances from the population mean in standard deviations units, where a <em>positive value</em> indicates that the value is <em>above</em> the mean and a <em>negative value</em> that the value is <em>below</em> it. A <em>standard normal distribution</em> has $\\ \mu = 0$ and $\\ \sigma = 1$ .

The formula for the <em>z-scores</em> is as follows

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Solving the question

Using all the previous information and using formula [1], we have

<em>x</em> = 6 hours per day (the raw score).

$\\ \mu = 4.5$ hours per day.

$\\ \sigma = 1.3$ hours per day.

Then (without using units)

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{6 - 4.5}{1.3}$

$\\ z = \frac{1.5}{1.3}$

$\\ z = 1.15384 \approx 1.15$

We round the value of <em>z</em> to two decimals since most standard normal tables only have two decimals for z.

We can observe that z = 1.15, and it tells us that the value is 1.15 standard deviations units above the mean.

With this value for <em>z</em>, we can consult the <em>cumulative standard normal table</em>, and for this z = 1.15, we have a cumulative probability of 0.8749. That is, this table gives us P(z<1.15).

We can describe the procedure of finding this probability in the next way: At the left of the table, we have z = 1.1; we can follow the first line on the table until we find 0.05. With these two values, we can determine the probability obtained above, P(z<1.15) = 0.8749.

Notice that the probability for the z-score, P(z<1.15), of the raw score, P(x<6) are practically the same, $\\ P(z$ . For an exact probability, we have to use a z-score = 1.15384 (without rounding), that is, $\\ P(z$ . However, the probability is approximated since we have to round z = 1.15384 to z = 1.15 because of the use of the table.

Therefore, "<em>the probability that the amount of time spent on leisure activities per day for a randomly selected adult from the population of interest is less than 6 hours per day</em>" is about 0.8749.

We can see this result in the graphs below. First, for P(x<6) in $\\ N(4.5, 1.3)$ (red area), and second, using the standard normal distribution ( $\\ N(0, 1)$ ), for P(z<1.15), which corresponds with the blue shaded area.

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