Answer:
Confidence Interval - 2.290 < S < 2.965
Step-by-step explanation:
Complete question
A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.
Solution
Given
n=50
x=23.33
s=2.6
Alpha = 1-0.80 = 0.20
X^2(a/2,n-1) = X^2(0.10, 49) = 63.17
sqrt(63.17) = 7.948
X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69
sqrt(37.69) = 6.139
s*sqrt(n-1) = 18.2
![\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }](https://tex.z-dn.net/?f=%5Cleq%20%5Csigma%20%5Cleq%20s%5Csqrt%7B%5Cfrac%7Bn-1%7D%7BX%5E2%20_%7B%28n-1%29%2C%201-%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%20%7D)
confidence interval:
(18.2/7.948) < S < (18.2/6.139)
2.290 < S < 2.965
Answer:
4
Step-by-step explanation:
the Fundamental Theorem of Algebra states that for any polynomial of degree n, there are n roots, some of which may be complex
The polynomial shown is of degree 4 ( highest exponent of x )
Hence the polynomial has 4 roots/ zeros
Answer:
230 degrees
Step-by-step explanation:
draw a horizontal line from x and use congruent rule to have a full 360 degree circle. You can now subtract 60 and 70 from 360 degrees leaving you with 230 degrees as x
Answer:
Top 5% is 5.84 milliters and the bottom 5% is 5.60 millimeters.
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 5.72, \sigma = 0.07](https://tex.z-dn.net/?f=%5Cmu%20%3D%205.72%2C%20%5Csigma%20%3D%200.07)
Top 5%:
X when Z has a pvalue of 0.95. So X when Z = 1.645
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.645 = \frac{X - 5.72}{0.07}](https://tex.z-dn.net/?f=1.645%20%3D%20%5Cfrac%7BX%20-%205.72%7D%7B0.07%7D)
![X - 5.72 = 1.645*0.07](https://tex.z-dn.net/?f=X%20-%205.72%20%3D%201.645%2A0.07)
![X = 5.84](https://tex.z-dn.net/?f=X%20%3D%205.84)
Bottom 5%:
X when Z has a pvalue of 0.05. So X when Z = -1.645
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-1.645 = \frac{X - 5.72}{0.07}](https://tex.z-dn.net/?f=-1.645%20%3D%20%5Cfrac%7BX%20-%205.72%7D%7B0.07%7D)
![X - 5.72 = -1.645*0.07](https://tex.z-dn.net/?f=X%20-%205.72%20%3D%20-1.645%2A0.07)
![X = 5.60](https://tex.z-dn.net/?f=X%20%3D%205.60)
Top 5% is 5.84 milliters and the bottom 5% is 5.60 millimeters.