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. a. Is it plausible that X is normally distributed?b. For a random sample of 50 such pairs, what is the (approximate) probability that the sample mean courtship time is between 100 min and 125 min?c. For a random sample of 50 such pairs, what is the (approximate) probability that the total courtship time exceeds 150hr?d. Could the probability requested in (b) be calculated from the given information if the sample size were 15 rather than 50? Explain.

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!!!

3x3 = 9 and 2x2 = 4

step-by-step explanation:

answer:      

step-by-step explanation: