* ASAP Need help as soon as possible do not understand The attached table lists the mean and standard deviation of two different normal distributions. In each case, a sample of 20 individuals, and a sample of 50 individuals were taken. For each sample, calculate the probability that the mean was less than the given value. Fill in the table - round to 2 decimals.

mean standard deviation value n-20 n=50
Pr(mean -5 5 -5.2
12 3 12.5

answer(isometric);

we solve the problem using the hardy-wineberg equation.

let us start with the homozygous recessive proportion of individuals with sickle cell anaemia,

we have the proportion as q² = 0.25.

∴ q = 0.5 (frequency of hbs in population)

further,

solving for the frequency of normal cell by using the fact that,

p + q = 1

∴ p = 1 - 0.5 = 1 - 0.5 = 0.5

now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous carriers of sickle cell anaemia,

∴ 2pq = 2 x 0.5 x 0.5 = 0.5

and the homozygous normal cell individuals proportion is given by the expression,

∴ p² = 0.5² = 0.25

hence, the carriers for sickle cell anaemia are in proportion, 0.5, or 50%.