Explain how knowing 1+7 helps you find the sum for 7+1
2 answers:
We can use the concept of commutative property of addition here. We know that the addition follow the commutative property.
The commutative property of addition is given by
It means, if we change the order, it will not affect the result.
Therefore, using above mentioned property, we can see that
1+7 = 7+1
Hence, if we know the value of 1+7, then we can easily find the value of 7+1 because both are same.
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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> , and <em>the population standard deviation,</em>
. In this case, the variable follows a normal distribution with parameters
hours per day and
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 and
.
The formula for the <em>z-scores</em> is as follows
[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).
hours per day.
hours per day.
Then (without using units)
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, . For an exact probability, we have to use a z-score = 1.15384 (without rounding), that is,
. 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 (red area), and second, using the standard normal distribution (
), for P(z<1.15), which corresponds with the blue shaded area.
Answer:
Daniel can read his data and refer to line as best line of fit and estimate an average per set of hours.
Step-by-step explanation:
A line of fit draws a solid conclusion to the average for the hours spent during the amount of indicated hours. We draw a line of fit central fit and aim similar centrality as that similar results of the mean (without working out the mean we can draw a line perpendicular to the number of mean, but in line of fit we go central to all the descending or cascading results to include all results but just using one line), with one further consideration and that is balance if anything sticks out from the norm ie) weather conditions including data, we suggest if there is nothing to weigh the line of fit to a balancing outcome that shows the opposite of kilometres walked (eg. extreme higher mileage within the hour/s) then it may just alter the line a fraction of how many treks he did, but not in data less than 30 entries. Have attached an example where they classify in economics something outside the norm is called a misfit. Daniel can read his data and refer to line as best line of fit and estimate an average per set of hours. Here on the attachment you can read any misfit info and use the line coordination perpendicular to guide the indifference, the attachment shows it is not really included in the best line of fit as other dominating balances have occurred and therefore we have a misfit, all whilst using best line of fit to balance everything fairly.
