Answer :
Approximately 15.87% (or 0.1587) of these batteries would be expected to survive beyond 1000 days.
Given that the life of an automotive battery is normally distributed with a mean of 900 days and a standard deviation of 35 days, we can use the properties of the normal distribution to determine the fraction of batteries that would survive beyond 1000 days.
First, we need to calculate the z-score, which measures the number of standard deviations a value is from the mean. The z-score can be calculated using the formula:
[tex]z = (x - μ) / σ[/tex]
where x is the value (1000 days), μ is the mean (900 days), and σ is the standard deviation (35 days).
Plugging in the values, we get:
[tex]z = (1000 - 900) / 35[/tex]
z = 2.857
Next, we look up the z-score in the standard normal distribution table (also known as the Z-table) to find the corresponding percentile. The Z-table provides the area under the normal curve to the left of a given z-score.
In this case, a z-score of 2.857 corresponds to a percentile of approximately 0.9971. This means that approximately 99.71% of the batteries would have a life expectancy of less than 1000 days.
To find the fraction of batteries that would survive beyond 1000 days, we subtract the percentile from 1:
1 - 0.9971 = 0.0029
Therefore, approximately 0.0029 (or 0.29%) of the batteries would be expected to survive beyond 1000 days.
Learn more about normal distribution here:
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