Answer:
3.84% probability that it has a low birth weight
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 3466, \sigma = 546](https://tex.z-dn.net/?f=%5Cmu%20%3D%203466%2C%20%5Csigma%20%3D%20546)
If we randomly select a baby, what is the probability that it has a low birth weight?
This is the pvalue of Z when X = 2500. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{2500 - 3466}{546}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B2500%20-%203466%7D%7B546%7D)
![Z = -1.77](https://tex.z-dn.net/?f=Z%20%3D%20-1.77)
has a pvalue of 0.0384
3.84% probability that it has a low birth weight
Answer:
20 ft.²
Step-by-step explanation:
You can find the area (A) by multiplying the length (l) and width (w) of a shape together: l×w=A
Plug in the measurements for length and width and solve
4×5=A
20=A
Answer:
45.3191
Step-by-step explanation:
0.0229 is 100 times smaller that 2.29 but 1979 is 100 times greater than 19.79. Since they cancel each other out, the answer remains the same.
Answer:
c
Step-by-step explanation:
idk but im think im sure