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.
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Answer:
The value on the In side: 2
The first value on the Out side: 87
The second value on the Out side: 93
Step-by-step explanation:
First, I started by finding the differences by each step on both sides. I noticed that they increased together. 45-27 is 18, and it mirrored on the other side. 27-9 is 18, and it mirrored. By this I mean that 81-63=18 as well as 63-45=18. So I followed this pattern 45-38=7, so 9-7=2, and that gives us our first answer. 51-45=6, so 81+6=87, gives us our second. 57-51=6, so 87+6=93, and this gives us our third.
The third one down
Because the best line of fit is a line that goes between the plotted points and it HAS to be a straight line and most of the points close to the line
Answer:
Step-by-step explanation:
Use the cruise ship room price table in problem 7 above to determine the percentile rank of $6,962.
Data from problem 7:
Seaquoia River Cruises offers a 7-day travel package. The prices vary based on the room class booked, as indicated in the following spreadsheet:
Use the cruise ship room price table in problem 7 above to -1
Hope this answer helps you :)
Have a great day
Mark brainliest
7(N) + 5 It's 7 times a number first, then add 5
