Find the indicated probability by using an appropriate normalmodel to approximat

Question

Find the indicated probability by using an appropriate normalmodel to approximate the binomial distribution. Information on a seed packet claims that the germination rateis 92%. Assuming this is true, what is the probability that atleast 95% of the 160 seeds in the packet will germinate? A) .3336 B) .1251 C) .0808 D) .0174 E) .2514 Find the indicated probability by using an appropriate normalmodel to approximate the binomial distribution. Information on a seed packet claims that the germination rateis 92%. Assuming this is true, what is the probability that atleast 95% of the 160 seeds in the packet will germinate? A) .3336 B) .1251 C) .0808 D) .0174 E) .2514

Explanation / Answer

The mean = probability of success * number of trials = .92 *160 = 147.2
Standard deviation = sqrt(probability of success * probabilityof failure (1- p. of success) * trials) = sqrt(160 * .92 * .08) = 3.43162.
z-score = (.95 * 160 (at least 95%) - 147.2)/standarddeviation (3.43162) = (152-147.2)/(3.43162) = 1.39876
We can look it up in the z-table using the value (1.40rounded), and we get .9192 falls under this value, so to get thevalue over, we just do 1- .9192 and get C .0808.

C

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