Information on a packet of seeds claims that the germination rate is 75%.Should you be surprised if 141 of the 150 seeds in the packet germinate?
A)You should not be surprised.A germination rate of 94.0% is only 3.22 standard deviations above the mean.
B)You should be very surprised.A germination rate of 94.0% is 5.37 standard deviations above the mean.
C)You should not be surprised.A germination rate of 94.0% is only 5.37 standard deviations above the mean.
D)You should be very surprised.A germination rate of 94.0% is 8.06 standard deviations above the mean.
E)You should be very surprised.A germination rate of 94.0% is 6.72 standard deviations above the mean.

Question

Quizplus · Accepted Answer

The expected number of seeds to germinate from a packet of 150 seeds with a germination rate of 75% would be 112.5 (150 * 0.75). The standard deviation for this binomial distribution can be calculated as sqrt(npq) = sqrt(150 * 0.75 * 0.25) = 5.79. Therefore, a germination rate of 94.0% (141 seeds) is (141 - 112.5)/5.79 = 4.91 standard deviations above the mean. A difference of 5.37 standard deviations (option B) is extreme and unlikely to occur by chance, indicating that you should be very surprised. Option A is incorrect as it cites the incorrect value of 3.22 standard deviations. Option C is incorrect as it also cites the incorrect value of 5.37 standard deviations. Option D is incorrect as it overstates the number of standard deviations. Option E is incorrect as it cites a value of 6.72 standard deviations that is not supported by the calculations.

Information on a Packet of Seeds Claims That the Germination