Question

Miraculin - a protein naturally produced in a rare tropical fruit - can convert a sour taste into a sweet taste. Consequently, miraculin has the potential to be an alternative low-calorie sweetener. In Plant Science (May 2010), 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) = .25 .

Answer 1

Answer:

a) P(x > 120) = 0.74537

b) P(100 < x < 110) = 0.46777

c) Value of a = 110.7

Step-by-step explanation:

Let x = amount of miraculin produced (measured in micro grams per gram of fresh weight)

We are given Mean, $\mu$ = 105.3  and  Standard Deviation, $\sigma$ = 8.0

Also, since x is normally distributed so;

                    Z = $\frac{x - \mu}{\sigma}$ follows N(0,1)

a) P(x > 100) = P( $\frac{x - \mu}{\sigma}$ > $\frac{100 - 105.3}{8.0}$ ) = P(Z > -0.6625) = P(Z < 0.66) = 0.74537

b) P(100 < x < 110) = P(x < 110) - P(x <= 100)

     P(x <= 100) = 1 - P(x > 100) = 1 - 0.74537 = 0.25463

     P(x < 110) = P( $\frac{x - \mu}{\sigma}$ < $\frac{110 - 105.3}{8.0}$ ) = P(Z < 0.59) = 0.72240

Hence, P(100 < x < 110) = 0.72240 - 0.25463 = 0.46777

c) Given expression is P(x < a ) = 0.25

                       ⇒ P( $\frac{x - \mu}{\sigma}$ < $\frac{a - 105.3}{8.0}$ ) = 0.25

                       ⇒ P(Z < $\frac{a - 105.3}{8.0}$ ) = 0.25

By seeing the Z % table we find that the value of z which have an are of 25% is 0.6745 i.e.

                     $\frac{a - 105.3}{8.0}$ = 0.6745

So, value of a = 0.6745*8 + 105.3 = 110.7 .