Question

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. ​

Answer 1

Answer:

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

b) $P(X <7) =0.00208$

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:

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

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

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

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

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

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

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

$P(X <7) =0.00208$

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