## In Pant Science (May 2010), a group of Japanese environmental scientists investigated the ability of a hybrid tomato plant to produce miracu

Question

In Pant Science (May 2010), a group of Japanese environmental scientists investigated the ability of a hybrid tomato plant to produce miraculin. (Miraculin is a protein naturally produced in a rare tropical fruit that can convert a sour taste into a sweet taste).For a particular generation of the tomato plant, the amount x of miraculin produced 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) P(100 < x < 110) c) Find the value a for which P(x < a) = .25

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1 month 2021-09-12T04:36:18+00:00 1 Answer 0

a) P(x > 120) = 0.03288

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

c) value of a for which P(x < a) = 0.25

a = 99.91

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 105.3

Standard deviation = σ = 8.0

a) P(x > 120)

To work this, we first need To convert 120 go standardized/normalized/z-scores.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x – μ)/σ = (120 – 105.3)/8.0 = 1.84

To determine P(x > 120) = P(z > 1.84)

We use the tables for these probabilities now.

P(x > 120) = P(z > 1.84) = 1 – P(z ≤ 1.84) = 1 – 0.96712 = 0.03288

b) P(100 < x < 110)

We normalize the two value in the inequality

For 100,

z = (x – μ)/σ = (100 – 105.3)/8.0 = -0.66

For 110,

z = (x – μ)/σ = (110 – 105.3)/8.0 = 0.59

To determine P(100 < x < 110) = P(-0.66 < z < 0.59)

We use the tables for these probabilities now.

P(100 < x < 110) = P(-0.66 < z < 0.59) = P(z < 0.59) – P(z < -0.66) = 0.72240 – 0.25463 = 0.46777

c) value of a for which P(x < a) = .25

Let the z-score of a be z’

z’ = (x – μ)/σ = (a – 105.3)/8.0

P(x < a) = P(z < z’) = 0.25

Using the table, z’ = – 0.674

-0.674 = (a – 105.3)/8

a – 105.3 = -5.392

a = 105.3 – 5.392 = 99.91

Hope this Helps!!!