Question

Tomato as a taste modifier. Miraculin is a protein naturally produced in a rare tropical fruit that can convert a sour taste into a sweet taste. Refer to the Plant Science (May 2010) investigation of the ability of a hybrid tomato plant to produce miraculin, Exercise 4.99 (p. 263). Recall that the amount x of miraculin produced in the plant had a mean of 105.3 micrograms per gram of fresh weight with a standard deviation of 8.0. Consider a random sample of $\mathbf{n}\mathbf{=}\mathbf{64}$hybrid tomato plants and let$\overline{\mathbf{x}}$ represent the sample mean amount of miraculin produced. Would you expect to observe a value of $\overline{\mathbf{X}}$ less than 103 micrograms per gram of fresh weight? Explain.

Short answer

The probability that it is observed a sample mean below $\overline{x}<103$micrograms if the mean and standard deviation of the miraculin are, respectively,$\mu =105.3$$\sigma =8$ is 0.0107. The probability is very small. This is because the researcher has observed an extremely rare event.

Step-by-step solution

Given information

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.

Calculating the probability

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

To find this probability, invoke the Central limit theorem.

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

${\mathit{\mu }}_{\mathbf{x}}\mathbf{=}\mathit{\mu }$and${\mathit{\sigma }}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$

Therefore,

${\mu }_{\overline{x}}=105.3$

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

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

Therefore, find desired probability as follows:

$\begin{array}{c}\mathrm{P}(\overline{\mathrm{x}}<103)\\ =\mathrm{P}\left(\frac{\overline{\mathrm{x}}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}<\frac{103-105.3}{1}\right)\\ =\mathrm{P}(\mathrm{z}<-2.3)\end{array}$

$=0.0107$

Therefore, the required probability is 0.0107.

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

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

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