What quadrant is the crab in?
2 answers:
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Answer:
4.72 hours/day
Step-by-step explanation:
Mean time spent watching TV (μ) = 2.8 hours a day
Standard deviation (σ) = 1.5 hours a day
The 90th percentile (upper 10%) of a normal distribution has an equivalent z-score of roughly z = 1.282. The minimum time spent watching TV, X, at the 90th percentile is:
On a typical day, you must watch at least 4.72 hours of TV to be in the upper 10%.
Answer:
(a) The probability that a randomly selected U.S. adult uses social media is 0.35.
(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.
(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.
Step-by-step explanation:
Denote the events as follows:
<em>X</em> = an US adult who does not uses social media.
<em>Y</em> = an US adult between the ages 18 and 29.
<em>Z</em> = an US adult between the ages 30 and above.
The information provided is:
P (X) = 0.35
P (Z) = 0.78
P (Y ∪ X') = 0.672
(a)
Compute the probability that a randomly selected U.S. adult uses social media as follows:
P (US adult uses social media (<em>X'</em><em>)</em>) = 1 - P (US adult so not use social media)
Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.
(b)
Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:
P (Adults between 18 - 29 (<em>Y</em>)) = 1 - P (Adults 30 or above)
Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.
(c)
Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:
P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')
Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.