Chapter 8: Q99E (page 452)

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 $P (x > 120)$ .
b. Find $P (100 < x < 110)$ .
c. Find the value a for which $P (x < a) = 0.25$ .

Short Answer

Expert verified

a. $P (x > 120) = 0.0329$

b. $P (100 < x < 110) = 0.4678$

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,

$μ = 105.3 a n d σ = 8$

$P (x > 120) = 1 - P (x < 120) = 1 - P (z < 1.8375) = 1 - 0.9671 = 0.0329 P (x > 120) = 0.0329$

Therefore, the required probability is 0.0329.

Probability calculation when P(100<x<110)

$P (100 < x < 110) = P (x < 110) - P (x < 100) = P (z < 0.5875) - P (z < 0.6625) = P (z < 0.5875) - P (1 - P (z < 0.6625)) = 0.7224 - (1 - 0.7454) = 0.4678 P (100 < x < 110) = 0.4678$

Therefore, the required probability is 0.4678.

Probability calculation when P(x<a)=0.25

$P (x < a) = 0.25 P (\frac{x - μ}{σ} < \frac{a - μ}{σ}) = 0.25 P (z < \frac{a - 105.3}{8}) = 0.25 Φ (\frac{a - 105.3}{8}) = 0.25 \frac{a - 105.3}{8} = Φ^{- 1} (0.25) a = 105.3 + 8 \times Φ^{- 1} (0.25) a = 105.3 + 8 \times (- 0.67449) a = 99.90408 a ~ 99.9$