Mathematics, 08.04.2020 03:27 Galycon12

Births are approximately Uniformly distributed between the 52 weeks of the year. They can be said to follow a Uniform distribution from 1 to 53 (a spread of 52 weeks). Round answers to 4 decimal places when possible a. The mean of this distribution is born at the exact moment that week 36 begins is Px 36) between weeks 8 and 45 is P(8 < ®?45)- The probability that a person will be born after week 16 is P(x > b. The standard deviation is c. The probability that a person will be d. The probability that a person will be born h. Find the minimum for the lower f. P(x>201 x 46). Find the 48th percentile quartile.

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Answer from: babyboo6745

(a) The mean of the distribution of births between the 52 weeks of a year is 27.

(b) The standard deviation of the distribution of births between the 52 weeks of a year is 15.01.

(d) The probability that a person will be born between weeks 8 and 45 is 0.7115.

(e) The probability that a person is born after week 16 is 0.7115.

Step-by-step explanation:

The random variable X can be defined as the births between the 52 weeks of the year.

The random variable X is uniformly distributed with parameters a = 1 to b = 53.

The probability distribution function of X is:

$f_{X}(x)=\left \{ {{\frac{1}{b-a};\ a$

(a)

Compute the mean of the Uniformly distributed random variable X as follows:

$E(X)=\frac{1}{2}(a+b)$

$=\frac{1}{2}\times (1+53)$

=27

Thus, the mean of the distribution of births between the 52 weeks of a year is 27.

(b)

Compute the standard deviation of the Uniformly distributed random variable X as follows:

$SD(X)=\sqrt{\frac{1}{12}\times (b-a)^{2}}$

$=\sqrt{\frac{1}{12}\times (53-1)^{2}}$

=15.01

Thus, the standard deviation of the distribution of births between the 52 weeks of a year is 15.01.

(c)

Compute the probability that a person is born exactly at the beginning of week 36 as follows:

Use continuity correction.

P (X = 36) = P (36 - 0.5 < X < 36 + 0.5)

= P (35.5 < X < 36.5)

$=\int\limits^{36.5}_{35.5}{\frac{1}{53-1}}}\, dx\\$

$=\frac{1}{52}\times \int\limits^{36.5}_{35.5}{1}\, dx$

$=\frac{36.5-35.5}{52}$

=0.0192

Thus, the probability that a person is born exactly at the beginning of week 36 is 0.0192.

(d)

Compute the probability that a person will be born between weeks 8 and 45 as follows:

P(8

$=\frac{1}{52}\times \int\limits^{45}_{8}{1}\, dx$

$=\frac{45-8}{52}$

=0.7115

Thus, the probability that a person will be born between weeks 8 and 45 is 0.7115.

(e)

Compute the probability that a person is born after week 16 as follows:

$P(X16)=\int\limits^{53}_{16}{\frac{1}{53-1}}\, dx$

$=\frac{1}{52}\times \int\limits^{53}_{16}{1}\, dx$

$=\frac{53-16}{52}$

=0.7115

Thus, the probability that a person is born after week 16 is 0.7115.

Answer from: Quest

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation:

Answer from: Quest

(6,1)(5,4)slope = (4 - 1) / (5 - 6) = 3/-1 = -3y = mx + bslope(m) = -3use either of ur , = 6 and y = 1now sub and find b, the y int1 = -3(6) + b1 = -18 + b1 + 18 = b19 = bso ur equation is : y = -3x + we need it in standard formy = -3x + 3x to both sides3x + y = 19 < == standard form

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