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

= 1.7

4 0

3 years ago

Pemdas please help explain too please

saul85 [17]

Parenthesis

Exponent

Multiplication

Division

Addition

Subtraction

(left to right)

4²-(3.1+6.4)+4.5

According to PEMDAS, we will do what is inside the parenthesis.

3.1+6.4 = 9.5

4²-9.5+4.5

Then we do figure out the exponent.

4² = (4)(4) = 16

16-9.5+4.5

Then we do subtraction then addition because PEMDAS also tells us we should do adding and subtracting from left to right. In this expression, subtracting is on the left side so we should do that first.

16-9.5 = 6.5

6.5 + 4.5 = 11

11 is your answer :)

6 0

3 years ago

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