One-fifth of the sum of q and 4

Mathematics

1 answer:

pashok25 [27]3 years ago

4 0

Answer:

.5/q+4

Step-by-step explanation:

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G = 1, h = 2 7 + g − 8 + 2h + 4g

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The weights of adult male birds of a certain species are normally distributed with a mean of 27.5 grams and a standard deviation

Korvikt [17]

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( $\mu=27.5,\sigma^{2} =1.1^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{ X-\mu}{\sigma } }$ ~ N(0,1)

where, $\mu$ = population mean weight = 27.5 grams

$\sigma$ = 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 = $\frac{ X-\mu}{\sigma } }$ = $\frac{ 29.37-27.5}{1.1 } }$