Mathematics, 13.03.2020 03:29 Jaedenaleinson

A social network provides a variety of statistics on its website that detail the growth and popularity of the site. On average, 21 percent of 18 to 34-year-olds check their social network profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. (a) Find the probability that the percent of 18 to 34-year-olds who check the social network before getting out of bed in the morning is at least 23. (Round your answer to four decimal places.)

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Answer from: morelos7078

The probability is 0.3446.

Step-by-step explanation:

We have here percentages that are normally distributed. In this case, we are talking about a group of people from 18 to 34-years-old "that check their social network profiles before getting out of bed in the morning".

The mean is 21 percent of them have this behavior (on average). The standard deviation is five percent. If the data is normally distributed, it implies that for 23 percent of them, there must be more cases for those that check their network profiles (for a cumulative probability).

The question is asking about the probability that the percent is at least 23. Since the mean is 21 percent, for normally distributed data, there is at least a probability of 0.50 that these people check their social network at this percentage, that is, 50% of the cases corresponds to that 21 percent of this people. What about 23 percent of people?

To find the probability in the question, we need to "transform" this raw score of 23 to a z-score to consult the cumulative standard normal table. We have to remember that with the standard normal distribution, we can obtain the probabilities associated with any normally distributed data and that the z-scores give us the distance from the mean in standard deviation units.

The formula for z-scores is as follows:

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

Where

$\\ x\;is\;the\;raw\;score$ .

$\\ \mu\;is\;the\;population\;mean$ .

$\\ \sigma\;is\;the\;population\;standard\;deviation$ .

The raw score here is 23 percent.

Then

$\\ z = \frac{23 - 21}{5}$

$\\ z =0.40$

The probability for a z = 0.40, using the cumulative standard normal table, we have that $\\ P(x$ (rounding to four decimal places).

So, there is a cumulative probability of 0.6554 for cases below that 23 percent (cumulative probability). However, for cases above 23 or at least 23 percent (or it could be more than 23), the corresponding probability is of P(x>23) = P(z>0.40) = 1 - 0.6554 = 0.3446 (or 34.46%) (i.e., the complement) that at least 23 percent of 18 to 34-year-olds "check the social network before getting out of bed in the morning".

See the graph below.