Question

Let X denote the courtship time for a randomly selected female-male pair of mating scorpion flies (time from the beginning of interaction until mating). Suppose the mean value of X is 120 min and the standard deviation of X is 110 min (suggested by data in the article "Should I Stay or Should I Go? Condition and Status-Dependent Courtship Decisions in the Scorpion Fly Panorpa Cognate"). (a) Is it plausible that X is normally distributed? No, courtship time cannot plausibly be normally distributed. Since X must be non-negative, realistically the interval + 3a should be entirely non negative which is not true in this case. Yes, courtship time is plausibly normally distributed. Since X must be non negative, realistically the interval 3 should be entirely non-negative which is true in this case. No, courtship time cannot plausibly be normally distributed. Since X must be negative, realistically the interval Jo should be entirely negative which is not true in this case Yes, courtship time is plausibly normally distributed. Since X must be negative, realistically the interval 30 should be entirely negative which is true in this case. Yes, courtship time is plausibly normally distributed since the mean of x is larger than 100. the approximate probability that the sample mean courtship time is between 115 min and 135 min? (Round your (b) For a random sample of 90 such pairs, what answer to four decimal places.) (e) For a random sample of places.) such pairs, what is the approximate) probability that the total courtship time exceeds 150 min? (Round your answer to four decimal

Answer 1

a)

otion e)

b)

Here, μ = 120, σ = 11.5950, x1 = 115 and x2 = 135. We need to compute P(115<= X <= 135). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (115 - 120)/11.5950 = -0.43
z2 = (135 - 120)/11.5950 = 1.29

Therefore, we get
P(115 <= X <= 135) = P((135 - 120)/11.5950) <= z <= (135 - 120)/11.5950)
= P(-0.43 <= z <= 1.29) = P(z <= 1.29) - P(z <= -0.43)
= 0.9015 - 0.3336
= 0.5679

c)

Here, μ = 120, σ = 11.595 and x = 150. We need to compute P(X >= 150). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (150 - 120)/11.595 = 2.59

Therefore,
P(X >= 150) = P(z <= (150 - 120)/11.595)
= P(z >= 2.59)
= 1 - 0.9952 = 0.0048

d)

OPtion B)
yes