Question

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In the following problem, check that it is appropriate to use the normal appreciation to the binomial. Then use the normal distribution to estimate the requested probabilities Do you try to padan insurance claim to cover your deductible? About 36% of all U.S. adults will try to pad their insurance dans Suppose that you are the director of an insurance adjustment office. Your office has just received 124 Insurance aims to be processed in the next few days. Find the following probabilities. (Round your answers to four decimal places.) (a) half or more of the dams have been added (b) fewer than 45 of the claims have been padded (c) from 40 to 64 of the claims have been padded (d) more than 10 of the claims have not been added

Answer 1

Given that about 36% of all U.S. adults will try to pad their insurance claims.

Hence p=probability that an U.S. adult will try to pad their insurance claims=0.36

Now the office received 124 insurance claims to be processed in the next few days.

Let X=Number of claims that are being padded.

$X\sim Binomial(n=124,p=0.36)$

What is Binomial Distribution?

A discrete random variable X is said to have a binomial distribution if its PMF(Probability Mass Function) is given by,

$f_{X}(x)=\left\{\begin{matrix} \binom{n}{p}p^{x}(1-p)^{n-x} &,x=0,1,2,...,n \\ 0&, otherwise \end{matrix}\right.$

Notation: X~Binomial(n,p)

Now here n is very large that makes our calculations hard and time consuming that is why we use normal approximation to the binomial distribution.

We can use normal approximation if np and np(1-p) are both greater than 5

$np=124*0.36=44.64> 5$

$np(1-p)=124*0.36*0.64=28.57=44.64> 5$

Hence we can use normal approximation to binomial.

Normal Approximation to the Binomial Distribution

If X~Binomial(n,p) then for large n,

$Z=\frac{X-np}{\sqrt{np(1-p)}}\sim Normal(0,1)$[Approximately for large n]

Normal Distribution

A continuous random variable X is said to have a normal distribution if its PDF(Probability Density Function) is given by

$f_{X}(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\left ( \frac{x-\mu }{\sigma } \right )^2},x\epsilon \mathbb{R}$

its CDF(Cumulative Distribution Function) is given by,

$P(X\leq x)=F(x)=\int_{-\infty }^{x}f_{X}(t)dt=\int_{-\infty }^{x}\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\left ( \frac{t-\mu }{\sigma } \right )^2}dt$

Notation:

$X\sim Normal(\mu ,\sigma ^{2})$

Standard Normal Distribution

A continuous random variable X is said to have a standard normal distribution if its PDF(Probability Density Function) is given by

$f_{X}(x)=\frac{1}{ \sqrt{2\pi}}e^{-\frac{1}{2}x^{2}},x\epsilon \mathbb{R}$

its CDF(Cumulative Distribution Function) is given by,

$P(X\leq x)=\Phi (x)=\int_{-\infty }^{x}f_{X}(t)dt=\int_{-\infty }^{x}\frac{1}{ \sqrt{2\pi}}e^{-\frac{1}{2}t^{2}}dt$

Exact evaluation of ?(x) is not possible but numerical method can be applied. The values of ?(x) has been tabulated extensively in Biometrika Volume I.

Notation:

$X\sim Normal(0,1)$

Continuity Correction Factor

Continuity correction is a correction that we use when a continuous distribution is used to approximate a discrete distribution.

$P(X=a)=P(a-0.5< X < a&plus;0.5 )$

Coming back to our problem,.

(a) Half or more of the plans have been padded,

$P\left ( X\geq \frac{124}{2} \right )$

$=P\left ( X\geq 62 \right )$

$=1-P(X<62)$

Applying continuity correction,

$=1-P(X<61.5)$

$=1-P\left ( \frac{X-np}{\sqrt{np(1-p)}}<\frac{61.5-np}{\sqrt{np(1-p)}} \right )$

$=1-P\left ( Z<\frac{61.5-np}{\sqrt{np(1-p)}} \right )[Z\sim Normal(0,1)]$

$=1-P\left ( Z<\frac{61.5-44.64}{\sqrt{28.5696}} \right )$

$=1-P\left ( Z<3.15 \right )$

$=1-\Phi (3.15)$

$=1-0.9992$

$=0.0008$

(b) Fewer than 45 of the claims have been padded

$P\left ( X< 45 \right )$

Applying continuity correction,

$=P\left ( X< 44.5 \right )$

$=P\left ( \frac{X-np}{\sqrt{np(1-p)}}<\frac{44.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left ( Z<\frac{44.5-np}{\sqrt{np(1-p)}} \right )[Z\sim Normal(0,1)]$

$=P\left ( Z<\frac{44.5-44.64}{\sqrt{28.5696}} \right )$

$=P\left ( Z<-0.03 \right )$

$=\Phi (-0.03)$

$=1-\Phi (0.03)$

$=1-0.5120=0.488$

(c) From 40 to 64 of the claims have been padded,

$P(40\leq X\leq 64)$

Applying continuity correction,

$=P(39.5< X< 64.5)$

$=P(X<64.5)-P(X<39.5)$

$=P\left ( \frac{X-np}{\sqrt{np(1-p)}}<\frac{64.5-np}{\sqrt{np(1-p)}} \right )-P\left ( \frac{X-np}{\sqrt{np(1-p)}}<\frac{39.5-np}{\sqrt{np(1-p)}} \right )$

$=P\left ( Z<\frac{64.5-np}{\sqrt{np(1-p)}} \right )-P\left ( Z<\frac{39.5-np}{\sqrt{np(1-p)}} \right )[Z~Normal(0,1)]$

$=P\left ( Z<\frac{64.5-44.64}{\sqrt{28.5696}} \right )-P\left ( Z<\frac{39.5-44.64}{\sqrt{28.5696}} \right )$

$=P\left ( Z<3.72 \right )-P\left ( Z<-0.96 \right )$

$=\Phi \left ( 3.72 \right )-\Phi \left ( -0.96 \right )$

$=\Phi \left ( 3.72 \right )-(1-\Phi \left ( 0.96 \right ))$

$=0.9999-(1-0.8315)$

$=0.9999-0.1685$

$=0.8314$

(d) More than 80 of the claims have been padded,

$P(X>80)$

$=1-P(X\leq 80)$

Applying continuity correction,

$=1-P(X< 80.5)$

$=1-P\left ( \frac{X-np}{\sqrt{np(1-p)}}<\frac{80.5-np}{\sqrt{np(1-p)}} \right )$

$=1-P\left ( Z<\frac{80.5-np}{\sqrt{np(1-p)}} \right )[Z\sim Normal(0,1)]$

$=1-P\left ( Z<\frac{80.5-44.64}{\sqrt{28.5696}} \right )$

$=1-P\left ( Z<6.71 \right )$

$=1-\Phi \left ( 6.71 \right )$

$\simeq 1-1\simeq 0$