Question

Tomato as a taste modifier. Miraculin—a protein naturally produced in a rare tropical fruit—has the potential to be an alternative low-calorie sweetener. In Plant Science (May2010), a group of Japanese environmental scientists investigated the ability of a hybrid tomato plant to produce miraculin. For a particular generation of the tomato plant, the amount x of miraculin produced (measured in micrograms per gram of fresh weight) had a mean of 105.3 and a standard deviation of 8.0. Assume that x is normally distributed.

a. Find$\mathit{P}\left(x>120\right)$.

b. Find$\mathit{P}\mathbf{(}\mathbf{100}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{110}\mathbf{)}$.

c. Find the value a for which$\mathit{P}\left(x<a\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}$.

Short answer

a.$P\left(x>120\right)=0.0329$

b.$P(100<\mathrm{x}<110)=0.4678$

c.$a=99.9$

Step-by-step solution

Given information

For a particular generation of the tomato plant, the amount x of miraculin produced had a mean of 105.3 and astandard deviation of 8.0.

Assume that x is normally distributed.

Probability calculation when P(x>120)

Here, the mean and standard deviation of the random variable x is given by,

$\mu =105.3and\sigma =8$

a.

$$\begin{gathered} P\left(x>120\right)=1-P\left(x<120\right) \\ =1-P\left(z<1.8375\right) \\ =1-0.9671 \\ =0.0329 \\ P\left(x>120\right)=0.0329 \end{gathered}$$

Therefore, the required probability is 0.0329.

Probability calculation when P(100<x<110)

b.

$$\begin{gathered} P\left(100<x<110\right)=P\left(x<110\right)-P\left(x<100\right) \\ =P\left(z<0.5875\right)-P\left(z<0.6625\right) \\ =P\left(z<0.5875\right)-P\left(1-P\left(z<0.6625\right)\right) \\ =0.7224-\left(1-0.7454\right) \\ =0.4678 \\ P\left(100<x<110\right)=0.4678 \end{gathered}$$

Therefore, the required probability is 0.4678.

Probability calculation when P(x<a)=0.25

c.

$$\begin{gathered} P\left(x<a\right)=0.25 \\ P\left(\frac{x-\mu }{\sigma }<\frac{a-\mu }{\sigma }\right)=0.25 \\ P\left(z<\frac{a-105.3}{8}\right)=0.25 \\ \Phi \left(\frac{a-105.3}{8}\right)=0.25 \\ \frac{a-105.3}{8}={\Phi }^{-1}\left(0.25\right) \\ a=105.3+8\times {\Phi }^{-1}\left(0.25\right) \\ a=105.3+8\times \left(-0.67449\right) \\ a=99.90408 \\ a~99.9 \end{gathered}$$

So, the required value of a is 99.9.