Question

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

Answer 1

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

Explanation:

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

The mean is, μ = 72 and the standard deviation is, σ = 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((X-\mu)/(\sigma)>(96.50-72)/(12))\\=P(Z>2.04)\\=1-P(Z<2.04)\\=1-0.97932\\=0.02068\\\approx 0.021`

*Use a z-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%.