How many groups of 1/3 are in 9
1 answer:
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Answer:
The probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers is 0.3156
Step-by-step explanation:
Mean output of amplifiers = 364
Standard deviation = = 12
We have to find the probability that the mean output for 52 randomly selected amplifiers will be greater than 364.8. Since the population is Normally Distributed and we know the value of population standard deviation, we will use the z-distribution to solve this problem.
We will convert 364.8 to its equivalent z-score and then finding the desired probability from the z-table. The formula to calculate the z-score is:
x=364.8 converted to z score for a sample size of n= 52 will be:
This means, the probability that the output is greater than 364.8 is equivalent to probability of z score being greater than 0.48.
i.e.
P( X > 364.8 ) = P( z > 0.48 )
From the z-table:
P( z > 0.48) = 1 - P(z < 0.48)
= 1 - 0.6844
= 0.3156
Since, P( X > 364.8 ) = P( z > 0.48 ), we can conclude that:
The probability that the mean amplifier output would be greater than 364.8 watts in a sample of 52 amplifiers is 0.3156