The gross income of Abelina Bennett is $215 per week. Her deductions are: $15.16, FICA tax; $29.33, income tax; 2% state tax; 1%
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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.
