Joel is getting a dog, a cat, and a hamster. He goes to the pet store and can choose from 8 dogs, 12 cats, and 5 hamsters. How m
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<em>The</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>X </em><em>is</em><em> </em><em>3</em><em>.</em><em>2</em><em>5</em>
<em>Look </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
Using the normal distribution, we have that:
a) The sketch of the situation is given at the end of this answer.
b) The probability is:
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
- It 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, which is the percentile of X.
In this problem:
- Mean of 4.1 years, thus
.
- Standard deviation of 1.3 years, thus
.
Item a:
The part between 2.8 and 7 years is shaded on the sketch given at the end of this answer.
Item b:
The probability is the <u>p-value of Z when X = 7 subtracted by the p-value of Z when X = 2.8</u>, thus:
X = 7:
has a p-value of 0.9871.
X = 2.8:
has a p-value of 0.1587.
0.9871 - 0.1587 = 0.8284, thus:
A similar problem is given at brainly.com/question/25151638
