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Mamont248 [21]
3 years ago
12

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

lly distributed with a mean of 4.5 hours per day and a standard deviation of 1.3 hours per day. Find 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. Round your answer to four decimal places. (make sure to put a 0 in front of the decimal ie 0.1 vs .1)

Mathematics
1 answer:
OLga [1]3 years ago
5 0

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:

  1. Use the <em>cumulative</em> <em>standard normal distribution</em> to find the probability.
  2. 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.
  3. 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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