Given thatx-3+ y and that -2y+3<-1, provide
1 answer:
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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.
Answer:
(-0.1059 ; - 0.0337)
Step-by-step explanation:
The data table is attached in the picture below:
These is a matched pair design ; which requires taking the difference of the two values for each sample :
The mean and standard deviation of the difference will be used to construct the confidence interval :
The mean of difference, dbar = Σx/n = - 0.0698
The standard deviation of difference, Sd ;
Sd = [√Σ(d - dbar)²/(n-1)] = 0.1054
n = sample size = 25
The confidence interval :
dbar ± [TCritical * Sd/√n]
Tcritical at 90% ; df = n -1 = 25 -1
Tcritical(90% , 24) = 1.1711
C.I = - 0.0698 ± (1.711 * 0.1054/√25)
C.I = - 0.0698 ± 0.0361
C.I = (-0.1059 ; - 0.0337)
