The admissions office at a university sends out 1000 admission letters to prospective students. The probability that an admitted student actually enrolled at that university is 0.4. Find the probability that fewer than 420 of the admitted students will enroll at this university.

The probability that fewer than 420 of the admitted students will enroll at this university is 0.9015

Step-by-step explanation:

The binomial distribution is a discrete distribution in which n trials can produce a success or a failure. We call the probability of success p and the probability of failure will be q = (1 - p).

A binomial distribution B (n, p) can be approximated by a normal distribution, provided that n is large and p is not very close to 0 or 1. The approximation consists of using a normal distribution with the same mean and standard deviation as the distribution binomial.

In practice, the approximation is used when n≥ 30, np ≥ 5 and n (1 - p) ≥ 5 and also p is close to 0.5.

The mean and standard deviation of the normal distribution are obtained by the expressions:

μ= n*p

σ= √n*p*q=√n*p*(1-p)

In this case, you know that n=1000 and p=0.4. Then:

μ= 1000*0.4= 400

σ= √1000*0.4*(1-0.4)= √240= 15.5

With X being the number of admitted students who enroll in this university and z=(X-μ)÷σ, then:

P(X≤420)=P(z≤(420-400)/15.5)= P(z≤ 1.29)= 0.9015

The probability that fewer than 420 of the admitted students will enroll at this university is 0.9015

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb