Respuesta :
Answer:
The z-score of a male bird of this species with a weight of 29.37 grams is 1.7.
Step-by-step explanation:
We are given that the weights of adult male birds of a certain species are normally distributed with a mean of 27.5 grams and a standard deviation of 1.1 grams.
Let X = weights of adult male birds of a certain species
So, X ~ Normal([tex]\mu=27.5,\sigma^{2} =1.1^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{ X-\mu}{\sigma } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean weight = 27.5 grams
[tex]\sigma[/tex] = standard deviation = 1.1 grams
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
SO, the z-score of a male bird of this species with a weight of 29.37 grams is given by;
Z score = [tex]\frac{ X-\mu}{\sigma } }[/tex] = [tex]\frac{ 29.37-27.5}{1.1 } }[/tex]
= 1.7
