Answer:
Probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.
Step-by-step explanation:
We are given that the diameters of ball bearings are distributed normally. The mean diameter is 106 millimeters and the standard deviation is 4 millimeters.
<em>Firstly, Let X = diameters of ball bearings</em>
The z score probability distribution for is given by;
Z =
~ N(0,1)
where,
= mean diameter = 106 millimeters
= standard deviation = 4 millimeter
Probability that the diameter of a selected bearing is greater than 111 millimeters is given by = P(X > 111 millimeters)
P(X > 111) = P(
>
) = P(Z > 1.25) = 1 - P(Z
1.25)
= 1 - 0.89435 = 0.1056
Therefore, probability that the diameter of a selected bearing is greater than 111 millimeters is 0.1056.
Answer:
10 gallons
Step-by-step explanation:
Amy's car's maximum capacity to hold gas = 19 gallons
present amount of gas = 9 gallons
As the car's capacity is 19 gallons and present amount in tank is 9 gallons, so we can fill it as much as so that the amount in tank becomes 19 gallons.
let assume she fills x gallons
then x gallons and 9 gallons present earlier should be equal to 19 gallon, as the tank can not hold more than that.
lets write it mathematically
x + 9 = 19
subtracting 9 from both side
x + 9 - 9 = 19 - 9
=> x = 10
Thus, 10 gallons of gas is needed to fill the gas tank.
Answer:
Both of these equations are in slope-intercept form. The slope-intercept form of a linear equation is: y=mx+b
Where m is the y-intercept value.
y=-2x+5
y=-2x+20
The slope of the two equations are: m1=−2 and m2=−2
Step-by-step explanation:
Because the have the same slope it means the lines represented by these two equations are either parallel or are the same line.
The y-intercepts for the two lines are:b1 = 5 and b1=20
Answer:
a) 98.522
b) 0.881
c) The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time.
Step-by-step explanation:
a.
As the mentioned in the given instruction the co-variance is first computed in excel by using only add/Sum, subtract, multiply, divide functions.
Marks y Time spent x y-ybar x-xbar (y-ybar)(x-xbar)
77 40 5.1 1.3 6.63
63 42 -8.9 3.3 -29.37
79 37 7.1 -1.7 -12.07
86 47 14.1 8.3 117.03
51 25 -20.9 -13.7 286.33
78 44 6.1 5.3 32.33
83 41 11.1 2.3 25.53
90 48 18.1 9.3 168.33
65 35 -6.9 -3.7 25.53
47 28 -24.9 -10.7 266.43
![Covariance=\frac{sum[(y-ybar)(x-xbar)]}{n-1}](https://tex.z-dn.net/?f=Covariance%3D%5Cfrac%7Bsum%5B%28y-ybar%29%28x-xbar%29%5D%7D%7Bn-1%7D)
Co-variance=886.7/(10-1)
Co-variance=886.7/9
Co-variance=98.5222
The co-variance computed using excel function COVARIANCE.S(B1:B11,A1:A11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted co-variance is 98.52222.
b)
The correlation coefficient is computed as
![Correlation coefficient=r=\frac{sum[(y-ybar)(x-xbar)]}{\sqrt{sum[(x-xbar)]^2sum[(y-ybar)]^2} }](https://tex.z-dn.net/?f=Correlation%20coefficient%3Dr%3D%5Cfrac%7Bsum%5B%28y-ybar%29%28x-xbar%29%5D%7D%7B%5Csqrt%7Bsum%5B%28x-xbar%29%5D%5E2sum%5B%28y-ybar%29%5D%5E2%7D%20%7D)
(y-ybar)^2 (x-xbar)^2
26.01 1.69
79.21 10.89
50.41 2.89
198.81 68.89
436.81 187.69
37.21 28.09
123.21 5.29
327.61 86.49
47.61 13.69
620.01 114.49
sum(y-ybar)^2=1946.9
sum(x-xbar)^2=520.1




The correlation coefficient computed using excel function CORREL(A1:A11,B1:B11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted correlation coefficient is 0.881.
c)
The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time. It means that as the study time increases the marks of student also increases and if the study time decreases the marks of student also decreases.
The excel file is attached on which all the related work is done.