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

Answer 1

Complete Question:

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?

Answer:

a) It is plausible that X is normally distributed

b) probability that the sample mean courtship time is between 100 min and 125 min is 0.5269

Step-by-step explanation:

a)X denotes the courtship time for the scorpion flies which indicates that is a real - valued random variable, and since normal distribution is a continuous probability distribution for a real valued random variable, it is plausible that X is normally distributed.

b) Probability that the sample mean courtship time is between 100 min and 125 min

$\mu = 120\\n = 50$

$P(x_{1} < \bar{X} < x_{2} ) = P(z_{2} < \frac{x_{2}- \mu }{SD} ) - P(z_{1} < \frac{x_{2}- \mu }{SD})$

$SD = \sqrt{\frac{\sigma^{2} }{n} } \\SD = \sqrt{\frac{110^{2} }{50} } \\SD = 15.56$

$P(100 < \bar{X} <125 ) = P(z_{2} < \frac{125- 120 }{15.56} ) - P(z_{1} < \frac{100- 120 }{15.56})\\P(100 < \bar{X} <125 ) = P(z_{2} < 0.32 ) - P(z_{1} < -1.29)$

From the probability distribution table:

$P(z_{2} < 0.32 ) = 0.6255\\ P(z_{1} < -1.29) = 0.0986$

$P(100 < \bar{X} <125 ) = 0.6255 - 0.0986\\P(100 < \bar{X} <125 ) =0.5269$