The average number of applications of one gallon of car wax is 450; with a stdev of 23.9; if the bottom 5.2% of the wax containers is defective from the factory; what is the lowest amount of applications will pass and not be rejected.
To find the lowest amount of applications that will not be rejected, we need to consider the standard deviation and the percentage of defective wax containers.
First, let's calculate the z-score for the bottom 5.2% of the distribution. The z-score will tell us how many standard deviations away from the mean this value lies.
Using a standard normal distribution table or a z-score calculator, we find that the z-score for the bottom 5.2% is approximately -1.66 (rounded to two decimal places).
The formula to calculate the lowest amount of applications that will not be rejected is:
Lowest amount = mean - (z-score * standard deviation)
Substituting the given values:
Lowest amount = 450 - (-1.66 * 23.9)
Now, let's solve for the lowest amount:
Lowest amount = 450 + (1.66 * 23.9)
Lowest amount = 450 + 39.604
Lowest amount = 489.604
Therefore, the lowest amount of applications that will pass and not be rejected is approximately 489.604 (rounded to three decimal places).