Answer:
The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.
Step-by-step explanation:
The confidence interval for population variance is given as below:
![[(n - 1)\times S^{2} / X^{2} \alpha/2, n-1 ] < \alpha < [(n- 1)\times S^{2} / X^{2} 1- \alpha/2, n- 1 ]](https://tex.z-dn.net/?f=%5B%28n%20-%201%29%5Ctimes%20S%5E%7B2%7D%20%20%2F%20%20X%5E%7B2%7D%20%20%5Calpha%2F2%2C%20n-1%20%5D%20%3C%20%5Calpha%20%3C%20%5B%28n-%201%29%5Ctimes%20S%5E%7B2%7D%20%20%2F%20X%5E%7B2%7D%201-%20%5Calpha%2F2%2C%20n-%201%20%5D)
We are given
Confidence level = 98%
Sample size = n = 81
Degrees of freedom = n – 1 = 80
Sample Variance = S^2 = 3.23
![X^{2}_{[\alpha/2, n - 1]} = 112.3288\\\X^{2} _{1 -\alpha/2,n- 1} = 53.5401](https://tex.z-dn.net/?f=X%5E%7B2%7D_%7B%5B%5Calpha%2F2%2C%20n%20-%201%5D%7D%20%20%20%3D%20112.3288%5C%5C%5CX%5E%7B2%7D%20_%7B1%20-%5Calpha%2F2%2Cn-%201%7D%20%3D%2053.5401)
(By using chi-square table)
[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 1 -α/2, n– 1 ]
[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]
2.3004 < σ^2 < 4.8263
Lower limit = 2.30
Upper limit = 4.83.
The correct answer is option (c) <span>P(not a hit) = 0.636
</span>The batting average of the baseball player =
= 0.364
Now, the probability of occurrence of any event is say, P, the probability of not occurrence of the event is given as = 1 - P
Therefore, <span>the probability that he will not get a hit at his next at bat = 1- 0.364 = 0.636</span>
Answer:
A. x1/15
Step-by-step explanation:
Answer:
Would be the 3rd one I believe
Step-by-step explanation:
Q1 of company A = 2.5
Q3 of company A = 8
Interquatile range = (Q3 - Q1)/2 = (8 - 2.5)/2 = 5.5/2 = 2.75
Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75
Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>
