I'm pretty sure it's just 2,420.
The distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.
<h3>Probability</h3>
a. Distribution
X ~ N (20 , 6)
b. P(x ≤24)
= P[(x - μ ) / σ (24 - 20) / 6]
= P(z ≤0.67)
= 0.74857
=0.7486
Hence:
Probability = 0.7486
c. P(21 < x < 26)
= P[(21 - 26)/ 6) < (x - μ ) / σ < (24 - 20) / 6) ]
= P(-0.83 < z < 0.67)
= P(z < 0.67) - P(z < -0.)
= 0.74857- 0.2033
= 0.54527
Hence:
Probability =0.54527
d. Using standard normal table ,
P(Z < z) = 66%
P(Z < 0.50) = 0.66
z = 0.50
Using z-score formula,
x = z× σ + μ
x = 0.50 × 6 + 20 = 23
23 Christmas cards
Therefore the distribution of X is X ~ N (20 , 6) and the probability that this American will receive no more than 24 Christmas cards this year is 0.7486.
Learn more about probability here:brainly.com/question/24756209
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