Daniels mom gave him $8 for his allowance He wants to buy new shoelaces for his vans that cost $5. What percentage of his money
1 answer:
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Answer:
Loss of $1.20
Step-by-step explanation:
The possible outcomes for this lottery and their probabilities are:
- a 1 in 100 chance of winning $450
- a 2 in 100 chance of winning $120
- a 4 in 100 chance of winning $30
- a 93 in 100 chance of losing $10
Therefore, the expected value of this lottery when buying one ticket is:
Therefore, you are expected to lose $1.2 per ticket.
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.