They are similar because they both contain variables.
Using the normal distribution, it is found that 95.65% of full term newborn female infants with a head circumference between 31 cm and 36 cm.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:

The proportion of full term newborn female infants with a head circumference between 31 cm and 36 cm is the <u>p-value of Z when X = 36 subtracted by the p-value of Z when X = 31</u>, hence:
X = 36:


Z = 1.83
Z = 1.83 has a p-value of 0.9664.
X = 31:


Z = -2.33
Z = -2.33 has a p-value of 0.0099.
0.9664 - 0.0099 = 0.9565.
0.9565 = 95.65% of full term newborn female infants with a head circumference between 31 cm and 36 cm.
More can be learned about the normal distribution at brainly.com/question/24537145
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Answer:
16/33
Step-by-step explanation:
We have a bag of :
9 red marbles
6 blue marbles
7 green marbles
11 yellow marbles
Total number of marbles = 33 marbles.
The possibility that a red or green marble will be selected from a bag
P( Red or Green) = P(Red) + P(Green)
In the question we are not told if it is with replacement or without. We do both
With replacement
P( R or G) = P(Red) + P(Green)
P(Red)= 9/33
P(Green) = 7/33
= 9/33 + 7/33
= 16/33
Therefore, the possibility that a red or green marble will be selected from a bag is 16/33
The simplified expression would be as follows: 5y + 26