The amount of daily time that teenagers spend on a brand A cell phone is normally distributed with a given mean mc027-1.jpg = 2.
5 hr and standard deviation mc027-2.jpg = 0.6 hr. What percentage of the teenagers spend more than 3.1 hr?
2 answers:
<span><span>If I read this properly, mean is 2.5 hr and sd is 0.6 hr.
z=(x-mean)/sd
=0.6/0.6, or 1
What is probability z>1?
It is 0.1587 or 16%. </span><span>
</span></span>
<span>16% is the answer Fam.....................................</span>
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