Question

question 6
2 pts
a report in 2010 indicates that americans between the ages of 8 and 18 spend an average of μ = 7.5 hours per day using some sort of electronic device such as smart phones, computers, or tablets. assume that the distribution of times is normal with a standard deviation of σ = 2.5 hours and find the following values.
what is the probability of selecting an individual who uses electronic devices more than 9 hours a day?
z = select ; p = select
what is the probability of selecting an individual who uses electronic devices less than 5 hours a day?
z = select ; p = select

Explanation:

Step1: Calculate z - score for x = 9

Use the z - score formula $z=\frac{x-\mu}{\sigma}$. Given $\mu = 7.5$, $\sigma=2.5$ and $x = 9$. Then $z=\frac{9 - 7.5}{2.5}=\frac{1.5}{2.5}=0.6$.

Step2: Find the probability for $z = 0.6$

We want $P(X>9)$, which is equivalent to $P(Z > 0.6)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z\leq z)$ can be found from the standard normal table. $P(Z\leq0.6)=0.7257$. So $P(Z > 0.6)=1 - P(Z\leq0.6)=1 - 0.7257 = 0.2743$.

Step3: Calculate z - score for x = 5

Use the z - score formula $z=\frac{x-\mu}{\sigma}$. Given $\mu = 7.5$, $\sigma = 2.5$ and $x = 5$. Then $z=\frac{5 - 7.5}{2.5}=\frac{-2.5}{2.5}=-1$.

Step4: Find the probability for $z=-1$

We want $P(X < 5)$, which is equivalent to $P(Z < - 1)$ in the standard normal distribution. From the standard normal table, $P(Z < - 1)=0.1587$.

Answer:

For $x = 9$: $z = 0.6$; $p=0.2743$
For $x = 5$: $z=-1$; $p = 0.1587$