Mathematics, 20.12.2019 03:31, makinzy03

Determine whether you can use a normal distribution to approximate the binomial distribution. if you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. if you cannot, explain why and use a binomial distribution to find the indicated probabilities. a survey of u. s. adults found that 64% watch nfl football. you randomly select 20 u. s. adults and ask them whether they watch nfl football. find the probability that the number who watch nfl football is

(a) exactly 10,
(b) less than 7, and
(c) at least 15, and
(d) identify any unusual events. explain.

determine whether a normal distribution can be used to approximate the binomial distribution.

choose the correct answer below.

a. yes, because both npgreater than or equals5 and nqgreater than or equals5. your answer is correct.
b. no, because npless than5.
c. no, because nqless than5.

Answers: 3

Show answers

Answers

answered: susanspagoot6385

a) P(X=10) = (20C10)(0.64)^{10} (1-0.64)^{20-10}=0.0779[/tex]

b) P(X

c) $P(X \geq 15) =0.217$

d) We can use the normal approximation since we check the conditions before. Unusual events would be given 3 deviations within the mean on this case

E(X) =np=20*0.64=12.8

$\sigma=\sqrt{np(1-p)}=\sqrt{20*0.64*(1-0.64)}=2.146$

12.8-3*(2.146)=6.36

12.8+3*(2.146)=19.239

So we can consider as unusual values of X less than 6 and higher than 19

Determine whether a normal distribution can be used to approximate the binomial distribution

$np=20*0.64=12.8 \geq 10$

n(1-p)=20*(1-0.64)=9.2

A. Yes, because both npgreater than or equals5 and nqgreater than or equals 5.

Step-by-step explanation:

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

Part a

$P(X=10) = (20C10)(0.64)^{10} (1-0.64)^{20-10}=0.0779$

Part b

P(X < 7) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)

And we can find the individual probabilities:

$P(X=0) = (20C0)(0.64)^{0} (1-0.64)^{20-0}=1.33x10^{-9}$

$P(X=1) = (20C1)(0.64)^{1} (1-0.64)^{20-1}=4.75x10^{-8}$

$P(X=2) = (20C2)(0.64)^{2} (1-0.64)^{20-2}=8.02x10^{-7}$

$P(X=3) = (20C3)(0.64)^{3} (1-0.64)^{20-3}=8.56x10^{-6}$

$P(X=4) = (20C4)(0.64)^{4} (1-0.64)^{20-4}=6.47x10^{-5}$

$P(X=5) = (20C5)(0.64)^{5} (1-0.64)^{20-5}=0.000368$

$P(X=6) = (20C6)(0.64)^{6} (1-0.64)^{20-6}=0.00164$

P(X

Part c

$P(X \geq 15) = P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)+P(X=20)$

$P(X=15) = (20C15)(0.64)^{15} (1-0.64)^{20-15}=0.116$

$P(X=16) = (20C16)(0.64)^{16} (1-0.64)^{20-16}=0.0644$

$P(X=17) = (20C17)(0.64)^{17} (1-0.64)^{20-17}=0.0270$

$P(X=18) = (20C18)(0.64)^{18} (1-0.64)^{20-18}=0.00799$

$P(X=19) = (20C19)(0.64)^{19} (1-0.64)^{20-19}=0.00150$

$P(X=20) = (20C20)(0.64)^{20} (1-0.64)^{20-20}=0.000133$

$P(X \geq 15) =0.217$

Determine whether a normal distribution can be used to approximate the binomial distribution

$np=20*0.64=12.8 \geq 10$

n(1-p)=20*(1-0.64)=9.2

A. Yes, because both npgreater than or equals5 and nqgreater than or equals 5.

Part d

We can use the normal approximation since we check the conditions before. Unusual events would be given 3 deviations within the mean on this case

E(X) =np=20*0.64=12.8

$\sigma=\sqrt{np(1-p)}=\sqrt{20*0.64*(1-0.64)}=2.146$

12.8-3*(2.146)=6.36

12.8+3*(2.146)=19.239

So we can consider as unusual values of X less than 6 and higher than 19

answered: Guest

introduction, methodology, results, discussion

step-by-step explanation:

an observational study is one which consists of actual observation to record data and then analyse the data.

there are a number of factors that should be analyzed before starting an observational study.

the main four factors are:

i) introduction: this should be analysed to check whether the purpose of the study is correctly established and the data recorded are appropriate for the study

methodology: the method of collecting the data, such as whether randomly selected, and whether sample size matches with the population to be studied, whether questions asked are easy to answer by common men, etc.

results: then results such as summary of data with comments should be studied thoroughly

discussion: the last is the discussion, this can be either by management people or persons who are all interested , etc.

answered: Guest

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation: