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A certain research center collected 20 nationally prepared food and compared the stated calorie contents per gram from the labels to calorie content determined in the laboratory.

The prior distribution is the normal distribution with mean 0 and a variance of 60.

Hence the priority density function is expressed a \(\xi \left( \theta \right) = 1\) for \(\theta \)

The likelihood function represents the relation between the posterior probability density function of\(\theta \)is proportional to the product of the likelihood function and the prior density function.

That is \(\xi \left( {\theta |x} \right)\alpha {f_n}\left( {x|\theta } \right)\xi \left( \theta \right)\)............(1)

Now, computing the posterior distribution of \(\theta \) for the prior improper is as follows.

The likelihood function is

\({f_n}\left( {x|\theta } \right)\alpha \exp \left( { - \frac{n}{{2\sigma }}{{\left( {\theta - \overline {{x_n}} } \right)}^2}} \right)\)

Also, we know that prior distribution is constant, and substituting the given values,

We get the equation (1) as

\(\begin{aligned}{}\xi \left( {\theta |x} \right) = \exp \left( { - \frac{n}{{2\sigma }}{{\left( {\theta - \overline {{x_n}} } \right)}^2}} \right) \times 1\\ = \exp \left( { - \frac{{20}}{{2 \times 60}}{{\left( {\theta - \left( { - 0.95} \right)} \right)}^2}} \right)\end{aligned}\)

As compared to the theorem, which states that the random variable \({X_1}...{X_n}\) forms a random sample from the normal distribution for which the value of the mean \(\theta \) unknown, but the value of the variance \({\sigma ^2}\) is known. Then the posterior distribution of \(\theta \) is given, that \({X_i} = {x_i}\left( {i = 1,...n} \right)\) is the normal distribution with mean \({\mu _1}\) and variance \(v_1^2\)

\({\mu _1} = \frac{{{\sigma ^2}{\mu _0} + nv_0^2\overline {{x_n}} }}{{{\sigma ^2} + nv_0^2}}\) and

\(v_1^2 = \frac{{{\sigma ^2}v_0^2}}{{{\sigma ^2} + nv_0^2}}\)

The function of \(\theta \) s is proportional to the probability density function of the normal distribution with a mean of -0.95 and

variance \(\frac{{60}}{{20}} = 3\)

let is compute the posterior probability that

\(\theta > 1\) is then

\(\begin{aligned}{}{\rm P}\left( {\theta > 1} \right) &= 1 - \phi \left( {\frac{{1 - \left( { - 0.95} \right)}}{{\sqrt 3 }}} \right)\\ &= 1 - \phi \left( {1.1258} \right)\\ &= 1 - 0.8699\end{aligned}\)

\({\rm P}\left( {\theta > 1} \right) = 0.1301\)

Therefore, the posterior probability \(\theta > 1\) is 0.1301