Leaving Cert Mathematics Statistics: A company produces calculator ba

A company produces calculator batteries - Leaving Cert Mathematics - Question 5 - 2012

Question 5

A company produces calculator batteries. The diameter of the batteries is supposed to be 20 mm. The tolerance is 0.25 mm. Any batteries outside this tolerance are re... show full transcript

Worked Solution & Example Answer:A company produces calculator batteries - Leaving Cert Mathematics - Question 5 - 2012

Step 1

The company has a machine that produces batteries with diameters that are normally distributed with mean 20 mm and standard deviation 0.1 mm. Out of every 10,000 batteries produced by this machine, how many, on average, are rejected?

96%

114 rated

Answer

To find the average number of rejected batteries, we need to calculate the probability that a battery falls outside the tolerance range of [19.75 mm, 20.25 mm].

First, we calculate the Z-scores for the two limits: $Z_{lower} = \frac{19.75 - 20}{0.1} = -2.5$ $Z_{upper} = \frac{20.25 - 20}{0.1} = 2.5$

Next, we find the probability that a battery falls outside these Z-scores: $P(|X - 20| > 0.25) = P(Z < -2.5) + P(Z > 2.5)$ Using standard normal distribution tables: $P(Z < -2.5) \approx 0.00621$ So, $P(Z > 2.5) \approx 0.00621$ Thus, $P(|X - 20| > 0.25) \approx 2 \times 0.00621 = 0.01242$

Now, out of 10,000 batteries, the average number rejected is: $10,000 \times 0.01242 \approx 124.2$ Therefore, approximately 124 batteries are rejected on average.

Step 2

A setting on the machine slips, so that the mean diameter of the batteries increases to 20.05 mm, while the standard deviation remains unchanged. Find the percentage increase in the rejection rate for batteries from this machine.

99%

104 rated

Answer

With the new mean of 20.05 mm and unchanged standard deviation of 0.1 mm, we recalculate the rejection probabilities.

We calculate new Z-scores: $Z_{lower} = \frac{19.75 - 20.05}{0.1} = -3.0$ $Z_{upper} = \frac{20.25 - 20.05}{0.1} = 2.0$

The probabilities are: $P(X < 19.75) = P(Z < -3.0) \approx 0.00135$ $P(X > 20.25) = P(Z > 2.0) \approx 0.02275$ Combining these gives: $P(X < 19.75) + P(X > 20.25) \approx 0.00135 + 0.02275 = 0.0241$

The new rejection rate is approximately 241 batteries out of 10,000.

To find the percentage increase:

Old rejection rate: 124 / 10,000 = 0.0124
New rejection rate: 241 / 10,000 = 0.0241

Percentage increase: $\text{Increase} = \frac{0.0241 - 0.0124}{0.0124} \times 100 = 94.35\%$ Thus, the rejection rate increases by approximately 94.35%.

A company produces calculator batteries - Leaving Cert Mathematics - Question 5 - 2012

Worked Solution & Example Answer:A company produces calculator batteries - Leaving Cert Mathematics - Question 5 - 2012

The company has a machine that produces batteries with diameters that are normally distributed with mean 20 mm and standard deviation 0.1 mm. Out of every 10,000 batteries produced by this machine, how many, on average, are rejected?

A setting on the machine slips, so that the mean diameter of the batteries increases to 20.05 mm, while the standard deviation remains unchanged. Find the percentage increase in the rejection rate for batteries from this machine.

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