Answer: 0.0129
Step-by-step explanation:
Given : Sample size : n=32
The average amount of time spent texting over a one-month period is : [tex]\mu=173\text{ minutes}[/tex]
Standard deviation : [tex]\sigma=66\text{ minutes}[/tex]
We assume that the time spent texting over a one-month period is normally distributed.
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 199
[tex]z=\dfrac{199-173}{\dfrac{66}{\sqrt{32}}}\approx2.23[/tex]
Now by using standard normal table, the probability that the average amount of time spent using text messages is more than 199 minutes will be :-
[tex]P(x>199)=P(z>2.23)=1-P(z\leq2.23)\\\\=1- 0.9871262=0.0128738\approx0.0129[/tex]
Hence, the required probability = 0.0129
