An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation of the distribution was 1.2 hours. About 95.44% of the batteries failed between what two values?
8.9 and 18.9
12.2 and 14.2
14.1 and 22.1
16.6 and 21.4
99.44% = mean ± 1.99 SD
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An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation of the distribution was 1.2 hours. About 95.44% of the batteries failed between what two values?
Selected Answer:
Correct 16.6 and 21.4
16.6 and 21.4
14.1 and 22.1
8.9 and 18.9
12.2 and 14.2
53. The mean score of a college entrance test is 500; the standard deviation is 75. The scores are normally distributed. What percent of the students scored below 320?
A) About 50.82%
B) About 34.13%
C) About 7.86%
D) About 0.82%
the averages starting salary of individuals with a master's degree in statistics is normally distributed with a mean of $48,000 and a standard deviation of $6,000. What is the probability that a randomly selected individual with a master's in statistics will get a starting salary of at least $55,500?
To find the range between which 95.44% of the batteries failed, we need to find the z-scores corresponding to the upper and lower limits of this percentage.
Step 1: Find the z-score corresponding to 97.72% (which is half of 95.44%).
The z-score corresponding to a percentage can be found using the z-table or a calculator. In this case, we need to find the z-score for the upper tail, so we need to subtract the percentage from 1.
The z-score corresponding to 97.72% can be calculated as:
z = InvNorm(0.9772) ≈ 2.00 (using a z-table or calculator)
Step 2: Calculate the raw scores (battery lives) corresponding to the z-scores.
The raw scores can be calculated using the formula:
x = μ + (z * σ)
where x is the raw score (battery life), μ is the mean, z is the z-score, and σ is the standard deviation.
Given that the mean (μ) is 19.0 and the standard deviation (σ) is 1.2, we can calculate the raw scores (battery lives) as follows:
For the lower limit:
xLower = 19.0 - (2.00 * 1.2) = 19.0 - 2.4 = 16.6
For the upper limit:
xUpper = 19.0 + (2.00 * 1.2) = 19.0 + 2.4 = 21.4
Therefore, about 95.44% of the batteries failed between the values of 16.6 and 21.4.
Therefore, the correct answer is:
16.6 and 21.4
Ball-Bearings, Inc. produces ball bearings automatically on a Kronar BBX machine. For one of the ball bearings, the mean diameter is set at 20.00 mm (millimeters). The standard deviation of the production over a long period of time was computed to be 0.150 mm. What percent of the ball bearings will have diameters of 20.27 mm or more?
Selected Answer:
Answers:
12.62%
3.59%
41.00%
85.00%