A scientist conducted a hybridization experiment using peas with green pods and

Question

A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed peas in such a way that 25% (or142142)of the568568offspring peas were expected to have yellow pods. Instead of getting142142peas with yellow pods, he obtained147147.Assume that the rate of 25% is correct.

a. Find the probability that among the568568offspring peas, exactly147147have yellow pods.

b. Find the probability that among the568568offspring peas, at least147147have yellow pods.

c. Which result is useful for determining whether the claimed rate of 25% is incorrect? (Part (a) or part (b)?)

d. Is there strong evidence to suggest that the rate of 25% is incorrect?

Explanation / Answer

NORMAL APPROXIMATION TO BINOMIAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 568 * 0.25 = 142
standard deviation ( npq )= 568*0.25*0.75 = 10.3199
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
P(X=147) = P(146.5 < X < 147.5)
BETWEEN THEM
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 146.5) = (146.5-142)/10.3199
= 4.5/10.3199 = 0.4361
= P ( Z <0.4361) From Standard Normal Table
= 0.6686
P(X < 147.5) = (147.5-142)/10.3199
= 5.5/10.3199 = 0.533
= P ( Z <0.533) From Standard Normal Table
= 0.70297
P(146.5 < X < 147.5) = 0.70297-0.6686 = 0.0344

b.

GREATER THAN EQUAL
P(X < 147) = (147-142)/10.3199
= 5/10.3199= 0.4845
= P ( Z <0.4845) From Standard NOrmal Table
= 0.686
P(X > = 147) = (1 - P(X < 147))
= 1 - 0.686 = 0.314

c.
Thc part (b) answer is more useful. In situations involving multiple order outcomes
a particular outcome is generally determined the probability of getting that outcome or
a more extreme chance it falls under
d.
No, since 0.314>0.05

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