91Ó°ÊÓ

The Given problem explains that the amount xof miraculin produced in the plant had a mean of 105.3 micro-gram of fresh weight with a standard deviation of 8.0.

A random sample of $n=64$hybrid tomato plants were considered, and the calculated sample mean.

Here, to find the probability of observing a value of$\stackrel{¯}{x}$less than 103 micrograms per gram of fresh weight. That is,$\mathrm{P}(\stackrel{¯}{\mathrm{x}}<103)$

To find this probability, invoke the Central limit theorem.

According to the theorem, the sampling distribution of $\stackrel{¯}{\mathit{x}}$has the following mean and standard deviation:

${\mathit{μ}}_{\mathbf{x}}\mathbf{=}\mathit{μ}$and${\mathit{σ}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{σ}}{\sqrt{\mathbf{n}}}$

Therefore,

${\mathrm{μ}}_{\stackrel{¯}{x}}=105.3$

$\begin{array}{c}{\mathrm{σ}}_{\mathrm{x}}=\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}\\ =\frac{8}{\begin{array}{l}\sqrt{64}\\ =1\end{array}}\end{array}$

The theorem also states that $\stackrel{¯}{x}$is approximately normally distributed.

Therefore, find desired probability as follows:

$\begin{array}{c}\mathrm{P}(\stackrel{¯}{\mathrm{x}}<103)\\ =\mathrm{P}\left(\frac{\stackrel{¯}{\mathrm{x}}−{\mathrm{μ}}_{\stackrel{¯}{\mathrm{x}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{x}}}}<\frac{103−105.3}{1}\right)\\ =\mathrm{P}(\mathrm{z}<−2.3)\end{array}$

$=0.0107$

Therefore, the required probability is 0.0107.

The probability that it is observed a sample mean below$\stackrel{¯}{\mathrm{x}}<103$

Micrograms, if the miraculin mean and standard deviation are, respectively, $\mathrm{μ}=105.3$and$\mathrm{σ}=8$ is 0.0107.

Since the researcher observed an extremely rare incident, the probability is incredibly low.