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?

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

Step-by-step 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:

*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%.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."