The length of a rectangle is 3 units greater than its width. If the width is w, what expression gives the perimeter of the recta
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Answer:
(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.
(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.
(c) Probability that a ball bearing is above the upper specification limit is 0.0401.
(d) Probability that a ball bearing is below the lower specification limit is 0.0006.
Step-by-step explanation:
We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.
Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.
Let X = <u><em>diameter of the ball bearings</em></u>
SO, X ~ Normal()
The z-score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 0.753 inch
= standard deviation = 0.004 inch
(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X 0.75 inch)
P(X < 0.753) = P( <
) = P(Z < 0) = 0.50
P(X 0.75) = P(
) = P(Z
-0.75) = 1 - P(Z < 0.75)
= 1 - 0.7734 = 0.2266
The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.
Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = <u>0.2734</u>.
(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X 0.74 inch)
P(X < 0.75) = P( <
) = P(Z < -0.75) = 1 - P(Z
0.75)
= 1 - 0.7734 = 0.2266
P(X 0.74) = P(
) = P(Z
-3.25) = 1 - P(Z < 3.25)
= 1 - 0.9994 = 0.0006
The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.
Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = <u>0.226</u>.
(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)
P(X > 0.76) = P( >
) = P(Z > -1.75) = 1 - P(Z
1.75)
= 1 - 0.95994 = <u>0.0401</u>
The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.
(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)
P(X < 0.74) = P( <
) = P(Z < -3.25) = 1 - P(Z
3.25)
= 1 - 0.9994 = <u>0.0006</u>
The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.
Answer:
Step-by-step explanation:
Recall that two events A,B are called mutually exclusive if and only if (their intersection is empty). They are exhaustive if they are mutually exclusive and their union is the sample space.
Based on this
a) Note that , so they are not mutually exclusive nor exhaustive.
b) so they are not mutually exclusive nor exhaustive.
c) , so they are mutually exclusive. Note that
. Then they are exhaustive.
d) , so they are not mutually exclusive nor exhaustive.