Answer:
(b) 0.214
Explanation:
Given S is normally distributed with µ=100 and σ=6, and T is normally distributed with µ=108 and σ=8, you want the probability that D = S-T is greater than 0. S and T are independent.
Difference
The distribution of the difference of the two normal random variables is normal with a mean equal to the difference of their means. The variance of the difference distribution will be the sum of the variances of the contributing distributions.
µ(D) = µ(S) -µ(T) = 100 -108 = -8
σ²(D) = σ²(S) +σ²(T) = 6² +8² = 100 ⇒ σ(D) = √100 = 10
Probability
The probability that the difference (d) will be greater than 0 can be found using a suitable probability calculator (see attached).
P(d > 0) ≈ 0.212
This is equivalent to the probability P(d > Z) for a standard normal distribution, where ...
Z = (d -µ)/σ = (0 -(-8))/10 = 0.8
The probability that Soledad sends more texts is about 0.214, choice B.
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Additional comment
We're not quite sure where the discrepancy between our value and the offered choice(s) comes from. It doesn't seem to be related to the empirical rule, which would apply for integer Z values. P(d > Z) = 0.214 will be true for Z ≈ 0.79262, not 0.8.
So, we have selected the closest choice.
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