Answer:
The score of a person who did better than 85% of all the test-takers was of 624.44.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score of a measure X is given by:
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 p-value, we get the probability that the value of the measure is greater than X.
One year, the average score on the Math SAT was 500 and the standard deviation was 120.
This means that ![\mu = 500, \sigma = 120](https://tex.z-dn.net/?f=%5Cmu%20%3D%20500%2C%20%5Csigma%20%3D%20120)
What was the score of a person who did better than 85% of all the test-takers?
The 85th percentile, which is X when Z has a p-value of 0.85, so X when Z = 1.037.
![X - 500 = 1.037*120](https://tex.z-dn.net/?f=X%20-%20500%20%3D%201.037%2A120)
![X = 624.44](https://tex.z-dn.net/?f=X%20%3D%20624.44)
The score of a person who did better than 85% of all the test-takers was of 624.44.
Answer:
a) ![\bar X = 369.62](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20369.62)
b) ![Median=175](https://tex.z-dn.net/?f=Median%3D175%20)
c) ![Mode =450](https://tex.z-dn.net/?f=%20Mode%20%3D450%20)
With a frequency of 4
d) ![MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5](https://tex.z-dn.net/?f=%20MidR%3D%20%5Cfrac%7BMax%20%2BMin%7D%7B2%7D%3D%20%5Cfrac%7B49%2B3000%7D%7B2%7D%3D%201524.5)
<u>e)</u>![s = 621.76](https://tex.z-dn.net/?f=%20s%20%3D%20621.76)
And we can find the limits without any outliers using two deviations from the mean and we got:
![\bar X+2\sigma = 369.62 +2*621.76 = 1361](https://tex.z-dn.net/?f=%20%5Cbar%20X%2B2%5Csigma%20%3D%20369.62%20%2B2%2A621.76%20%3D%201361)
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
![\bar X = \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
Replacing we got:
![\bar X = 369.62](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20369.62)
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
![Median=175](https://tex.z-dn.net/?f=Median%3D175%20)
Part c
The mode is the most repeated value in the sample and for this case is:
![Mode =450](https://tex.z-dn.net/?f=%20Mode%20%3D450%20)
With a frequency of 4
Part d
The midrange for this case is defined as:
![MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5](https://tex.z-dn.net/?f=%20MidR%3D%20%5Cfrac%7BMax%20%2BMin%7D%7B2%7D%3D%20%5Cfrac%7B49%2B3000%7D%7B2%7D%3D%201524.5)
Part e
For this case we can calculate the deviation given by:
![s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%20%3D%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
And replacing we got:
![s = 621.76](https://tex.z-dn.net/?f=%20s%20%3D%20621.76)
And we can find the limits without any outliers using two deviations from the mean and we got:
![\bar X+2\sigma = 369.62 +2*621.76 = 1361](https://tex.z-dn.net/?f=%20%5Cbar%20X%2B2%5Csigma%20%3D%20369.62%20%2B2%2A621.76%20%3D%201361)
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Answer:
![y=2(\frac{1}{2} )^x](https://tex.z-dn.net/?f=y%3D2%28%5Cfrac%7B1%7D%7B2%7D%20%29%5Ex)
Step-by-step explanation:
Given the exponential function as
![y=ab^x](https://tex.z-dn.net/?f=y%3Dab%5Ex)
substitute both points in the equation above
point (-3,16) will be
![16=ab^{-3}](https://tex.z-dn.net/?f=16%3Dab%5E%7B-3%7D)
point (-1,4) will be
![4=ab^{-1}](https://tex.z-dn.net/?f=4%3Dab%5E%7B-1%7D)
make a the subject of the formula in both equations above
![a=\frac{16}{b^{-3} } \\\\\\a=\frac{4}{b^{-1} }](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B16%7D%7Bb%5E%7B-3%7D%20%7D%20%5C%5C%5C%5C%5C%5Ca%3D%5Cfrac%7B4%7D%7Bb%5E%7B-1%7D%20%7D)
This means
![=\frac{16}{b^{-3} } =\frac{4}{b^{-1} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B16%7D%7Bb%5E%7B-3%7D%20%7D%20%3D%5Cfrac%7B4%7D%7Bb%5E%7B-1%7D%20%7D)
Cross multiply as;
![16b^{-1} =4b^{-3}](https://tex.z-dn.net/?f=16b%5E%7B-1%7D%20%3D4b%5E%7B-3%7D)
Divide by 4 both sides to get
![4b^{-1} =b^{-3}](https://tex.z-dn.net/?f=4b%5E%7B-1%7D%20%3Db%5E%7B-3%7D)
Divide by b^-1 both sides
![\frac{4b^{-1} }{b^{-1} } =\frac{4b^{-3} }{b^{-1} } \\\\\\4=b^{-2} \\\\\\4=\frac{1}{b^2} \\\\\\b^2=\frac{1}{4} \\\\\\b=\sqrt{\frac{1}{4} } =\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B4b%5E%7B-1%7D%20%7D%7Bb%5E%7B-1%7D%20%7D%20%3D%5Cfrac%7B4b%5E%7B-3%7D%20%7D%7Bb%5E%7B-1%7D%20%7D%20%5C%5C%5C%5C%5C%5C4%3Db%5E%7B-2%7D%20%5C%5C%5C%5C%5C%5C4%3D%5Cfrac%7B1%7D%7Bb%5E2%7D%20%5C%5C%5C%5C%5C%5Cb%5E2%3D%5Cfrac%7B1%7D%7B4%7D%20%5C%5C%5C%5C%5C%5Cb%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B4%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B2%7D)
Find value of a
![4=ab^{-1} \\\\4=a*(\frac{1}{2})^{-1} \\\\\\4=a*2\\\\2=a](https://tex.z-dn.net/?f=4%3Dab%5E%7B-1%7D%20%5C%5C%5C%5C4%3Da%2A%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B-1%7D%20%20%5C%5C%5C%5C%5C%5C4%3Da%2A2%5C%5C%5C%5C2%3Da)
Hence
a=2 and b=1/2 thus write the equation as;
![y=ab^x\\\\\\y=2(\frac{1}{2} )^x](https://tex.z-dn.net/?f=y%3Dab%5Ex%5C%5C%5C%5C%5C%5Cy%3D2%28%5Cfrac%7B1%7D%7B2%7D%20%29%5Ex)