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kipiarov [429]
3 years ago
8

A researcher was testing the number of popcorn kernels that popped out of a mini bag of 100 kernels after being cooked for the s

pecific time in the microwave. the researcher found that on average 72 currently parked with a standard deviation of 12.what percentage of the bag should have popped 96 kernels or more?
Mathematics
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

The percentage of the bag that should have popped 96 kernels or more is 2.1%.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of popcorn kernels that popped out of a mini bag.

The mean is, <em>μ</em> = 72 and the standard deviation is, <em>σ</em> = 12.

Assume that the population of the number of popcorn kernels that popped out of a mini bag follows a Normal distribution.

Compute the probability that a bag popped 96 kernels or more as follows:

Apply continuity correction:

P( X\geq 96)=P( X>96+0.50)

                 =P( X>96.50)\\=P(\frac{X-\mu}{\sigma}>\frac{96.50-72}{12})\\=P(Z>2.04)\\=1-P(Z

*Use a <em>z</em>-table.

The probability that a bag popped 96 kernels or more is 0.021.

The percentage is, 0.021 × 100 = 2.1%.

Thus, the percentage of the bag that should have popped 96 kernels or more is 2.1%.

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