When Mendel conducted his famous hybridization experiments, he used peas with gr

Question

When Mendel conducted his famous hybridization experiments, he used peas with green pods and yellow pods. One experiment involved crossing peas in such a way that 25% (or 145) of the 580 offspring peas were expected to have yellow pods. Instead of getting 145 peas with yellow pods, he obtained 152. Assume that Mendel's 25% rate is correct (or that the true proportion is 25%). Find the probability that among the 580 offspring peas, at least 152 have yellow pods. (Use the normal approximation if appropriate, otherwise, choose none of the above.)

a) 0.2514

b) 0.3167

c) 0.6833

Explanation / Answer

p is equal to .25 since 25%, or 145 out of 580 of offspring yield yellow peas.
q is equal to .75 since p + q = 1
n = 580 (total offspring)

the mean is equal to n*p which is 145 (also given in the problem)

the standard deviation is equal to square root of n*p*q.
SD= 10.4283

The z-score is found by (x-value - mean) / (SD) = (151.99 - 145) / (10.4283) =0.67029

After locating 0.67029 on the normal curve z-score chart, the probability is 0.748664

Since the question asked for AT LEAST that is the reason that:
Instead of 152 you use 151.99 because that would be including the integer 152
P(atleast 152 have yellow pods)=1 - 0.748664 = 0.251336 = 0.2514

Option (a) is correct

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