Question

1. Assume that a scientist finds that the distance of locust swarms in the second wave in Africa follows a normal distribution, with a mean of 1,800 kilometers and a standard deviation of 350 kilometers.

   1. What is the probability that a locust travels less than 2000 kilometers?

   2. Find a distance such that 95% of all locust travel less than this distance.

   3. What is the probability that a group of 100 locusts travel an average of more than 1880 km?

Answer 1

a)

Here, μ = 1800, σ = 350 and x = 2000. We need to compute P(X <= 2000). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (2000 - 1800)/350 = 0.57

Therefore,
P(X <= 2000) = P(z <= (2000 - 1800)/350)
= P(z <= 0.57)
= 0.7157

b)

z value at 95% = -1.64

z = (x - mean)/s
-1.64= (x - 1800)/350
x = -1.64 * 350 + 1800
x = 1226

c)
Here, μ = 1800, σ = 35 and x = 1880. We need to compute P(X >= 1880). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (1880 - 1800)/35 = 2.29

Therefore,
P(X >= 1880) = P(z <= (1880 - 1800)/35)
= P(z >= 2.29)
= 1 - 0.989 = 0.0110