Answer:
a) Check Explanation
b) P(100 < x < 125) = 0.52699
c) P(x > 150) = 0.0268
d) Check Explanation
Step-by-step explanation:
It seems implausible that this type of distribution is a normal distribution, but it really is not totally implausible that this distribution is a normal distribution because as a skewed distribution which the distribution is, it can also 'approximate' normal distributions.
It means the distribution of courtship time for a randomly selected female-male pair of mating scorpion flies (time from the beginning of interaction until mating) can vary normally from 0 through the mean in a normal skewed manner and vary normally after the mean.
But, the distribution of sample means for this 'approximate' normal-skewed distribution has been proven to approximate a normal distribution, even more when n > 30.
b) Population mean = μ = 120 min
Population Standard deviation = σ = 110 min
sample size = n = 50
Sample mean = μₓ = μ = 120 min
Standard deviation of the distribution of sample means = σₓ = (σ/√n) = (110/√50)
σₓ = 15.56 min
Approximate probability that the sample mean courtship time is between 100 min and 125 min = P(100 < x < 125)
We first need to convert 100 min and 125 min to standard z-scores.
The z-score for any value is the value minus the mean then divided by the standard deviation.
For 100 mins
z = (x - μ)/σ = (100 - 120)/15.56 = - 1.29
For 125 mins
z = (x - μ)/σ = (125 - 120)/15.56 = 0.32
To determine the approximate probability that the sample mean courtship time is between 100 min and 125 min
P(100 < x < 125) = P(-1.29 < z < 0.32)
We'll use data from the normal probability table for these probabilities
P(100 < x < 125) = P(-1.29 < z < 0.32)
= P(z < 0.32) - P(z < -1.29)
= 0.62552 - 0.09853 = 0.52699
P(100 < x < 125) = P(-1.29 < z < 0.32) = 0.52699
c) Approximate probability that the total courtship time exceeds 150 min = P(x > 150)
Converting 150 min into z-scores
z = (x - μ)/σ = (150 - 120)/15.56 = 1.93
To determine the approximate probability that the total courtship time exceeds 150 min
P(x > 150) = P(z > 1.93)
Using the normal distribution table
P(x > 150) = P(z > 1.93) = 1 - P(z ≤ 1.93)
= 1 - 0.9732
= 0.0268
d) Yes, the approximate probabilities could still be calculated, but they would be farther from the real probabilities, the smaller the value of n because the distribution of sample means approximates normal distributions more, as the sample size increases.
Hope this Helps!!!
