The number of ants per acre in the forest is normally distributed with mean 42,000 and standard deviation 12,236. Let X = number of ants in a randomly selected acre of the forest. Round all answers to 4 decimal places where possible. a. What is the distribution of X? X ~ N( , ) b. Find the probability that a randomly selected acre in the forest has fewer than 55,486 ants. c. Find the probability that a randomly selected acre has between 42,894 and 57,677 ants. d. Find the first quartile. ants (round your answer to a whole number)
Solution :
(a)
X ~ N(42000 , 122362)
(b)
P(x < 55486) = P[(x - ) / < (55486 - 42000) / 12236]
= P(z < 1.10)
= 0.8643
(c)
P(42894 < x < 57677)
= P[(42894 - 42000) / 12236) < (x - ) / < (57677 - 42000) / 12236) ]
= P(0.07 < z < 1.28)
= P(z < 1.28) - P(z < 0.07)
= 0.3718
(d)
P(Z < -0.674) = 0.25
z = -0.674
Using z-score formula,
x = z * +
x = -0.674 * 12236 + 42000 = 33753
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