Question

The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds produced from the plant was analyzed for a par- ticular collagen. The collagen amount was found to be normally distributed with a mean of 74 and standard deviation of 8.2 grams per milliliter.

(a)

[3 pts] What is the probability that the amount of collagen is greater than 60 grams per milliliter?

(b)

[3 pts] What is the probability that the amount of collagen is less than 78 grams per milliliter?

(c)

[3 pts] What percentage of compounds formed from the extract of this plant fall within 2 standard deviations of the mean?

Answer #1

Solution :

Given that ,

mean = = 74

standard deviation = = 8.2

a)

P(x > 60) = 1 - P(x < 60)

= 1 - P((x - ) / < (60 - 74) / 8.2)

= 1 - P(z < -1.71)

= 1 - 0.0436 Using standard normal table.

= 0.9564

**Probability = 0.9564**

b)

P(x < 78) = P((x - ) / < (78 - 74) / 8.2)

= P(z < 0.49)

= 0.6879 Using standard normal table,

**Probability = 0.6879**

c)

By using empirical rule,

- 2 = 74 - 2*82 = 57.6

and

+ 2 = 74 + 2*8.2 = 90.4

P( Between57.6 and 90.4 ) = P( - 2 and + 2 )

**= 95%**

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