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Suppose the diameter at breast height (in.) of a certain type of tree is distributed normally with a mean of 8.8 and a standard deviation of 2.8.

(a)What is the probability that the diameter of a randomly selected tree will be at least 10 inches? Will exceed 10 inches.

(b) What is the value of c so that the interval (8.8-c, 8.8+c) contains 98% of all diameter values.

Sagot :

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Answer:

a. 7% b. 1.4 inches

Step-by-step explanation:

(a)What is the probability that the diameter of a randomly selected tree will be at least 10 inches? Will exceed 10 inches.

Since our mean, x = 8.8 and standard deviation, σ = 2.8, and we want our maximum value to be 10 inches,

So, x + ε = 10 where ε = error between mean and maximum value

So,8.8 + ε = 10

ε = 10 - 8.8 = 1.2

Since σ = 2.8, ε/σ = 1.2/2.8 = 0.43

ε/σ × 100 % = 0.43 × 100 = 43%

Since 50% of our values are in the range x - 3σ to x and 43% of our values are in the range x to x + ε = x + 0.43σ, the probability of finding a value less than 10 inches is thus 50 % + 43% = 93%.

So, the probability of finding a value greater than 10 inches is thus 100 % -   93 % = 7 %.

(b) What is the value of c so that the interval (8.8-c, 8.8+c) contains 98% of all diameter values.

Since 98% of the values range from 8.8-c to 8.8+c, then half of the interval is from 8,8 - c to 8.8 or 8.8 to 8.8 + c. So, the number that range in this half interval is 98%/2 = 49%

So, c/σ × 100 % = 49%  

c/σ = 0.49

c = 0.49σ = 0.49 × 2.8 = 1.372 ≅ 1.37 inches = 1.4 inches to 1 d.p

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