The probability that a live birth requires a cesarean delivery in a certain American state is 0.248. If 8 live births occur inde
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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:
*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%.