5. A soft drink dispensing machine uses paper cups that hold a maximum of 12 ounces. The machine is set to dispense a mean of 10 ounces of the drink. The amount that is actually dispensed varies. It is normally distributed with a standard deviation of one ounce.
(a) What fraction of the drinks are between 10 and 11 ounces?
(b)What percentage of the cups spill over when poured because they exceed the 12 ounce limit?
Z = (score-mean)/SD = (score-10)/1
Insert scores to find Z scores.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.
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To solve this problem, we can use the properties of the normal distribution.
(a) To find the fraction of drinks between 10 and 11 ounces, we need to calculate the area under the normal curve between those two values.
First, we need to standardize the values using the formula: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation.
For x = 10 ounces:
z = (10 - 10) / 1 = 0
For x = 11 ounces:
z = (11 - 10) / 1 = 1
Next, we can use a Z-table or calculator to find the corresponding probabilities.
From the Z-table, the area to the left of z = 0 is 0.5000, and the area to the left of z = 1 is 0.8413.
Therefore, the fraction of drinks between 10 and 11 ounces is: 0.8413 - 0.5000 = 0.3413, or approximately 34.13%.
(b) To find the percentage of cups that spill over, we need to calculate the area under the normal curve above 12 ounces (since that is the maximum limit).
First, we need to standardize the value of 12 ounces:
z = (12 - 10) / 1 = 2
From the Z-table, the area to the left of z = 2 is 0.9772.
Therefore, the percentage of cups that spill over because they exceed the 12 ounce limit is: (1 - 0.9772) * 100 = 2.28%, or approximately 2.28%.
To calculate the fraction of drinks between 10 and 11 ounces, we need to find the area under the normal distribution curve between these two values. We can use the standardized Z-score formula to convert these values to z-scores and then use a standard normal distribution table or a statistical calculator to find the corresponding probabilities.
(a) Calculating the fraction of drinks between 10 and 11 ounces:
1. Convert the values to z-scores using the formula: z = (x - mean) / standard deviation.
For 10 ounces: z1 = (10 - 10) / 1 = 0
For 11 ounces: z2 = (11 - 10) / 1 = 1
2. Look up the corresponding probabilities from the standard normal distribution table or use a statistical calculator. The table or calculator will provide the area under the curve up to the given z-scores.
Using a standard normal distribution table, the values are as follows:
For z = 0, the area is 0.5000
For z = 1, the area is 0.8413
3. Subtract the smaller probability from the larger probability to find the fraction between 10 and 11 ounces:
Fraction = 0.8413 - 0.5000 = 0.3413 (approximately)
Hence, approximately 34.13% of the drinks are between 10 and 11 ounces.
(b) To calculate the percentage of cups that spill over because they exceed the 12-ounce limit, we need to find the probability that the amount dispensed is greater than 12 ounces.
1. Convert the value (12 ounces) to a z-score using the same formula:
z = (x - mean) / standard deviation
z = (12 - 10) / 1 = 2
2. Look up the corresponding probability from the standard normal distribution table or use a statistical calculator. A z-score of 2 corresponds to an area of 0.9772.
3. The percentage of cups that spill over is equal to:
Percentage = (1 - probability) * 100
Percentage = (1-0.9772) * 100 = 2.28% (approximately)
Hence, approximately 2.28% of the cups spill over when poured because they exceed the 12-ounce limit.