A certain type of elephant averages 6.4 tons with an SD of 1.2 tons. Assume that

Question

A certain type of elephant averages 6.4 tons with an SD of 1.2 tons. Assume that the amount would be normally distributed. A. What is the probability that a randomly selected elephant weights more than 7 tons? (enter a, b, c, d or e) a. 0.16 b. 0.31 c. 0.76 d. 0.24 e. 0.69 B. What is the probability that a randomly selected elephant weights between 8.2 to 9.4 tons? (enter a, b, c, d or e) a. 0.16 b. 0.31 c. 0.76 d.0.24 e. 0.06 C. How many of the next 100 elephants will weight less than 8.8 tons? (enter a, b, c, d or e) a. 43 b. 57 c.93 d. 97 e. 3

Explanation / Answer

1) Since =6.4 and =1.2 we have:

P ( X>7 )=P ( X>76.4 )

P(x>7)=P((X)/>(76.4)/1.2)

Since Z=(x)/ and( (76.4)/1.2)=0.5 we have:

P ( X>7 )=P ( Z>0.5 )

P (Z>0.5)=0.3085

The answer is"b"( 0.31)

2) =6.4 and =1.2 we have:

P ( 8.2<X<9.4 )=P ( 8.26.4< X<9.46.4 )

P(8.2<x<9.4) =P {( 8.26.4)/1.2<(X)/<(9.46.4)/1.2)}

Since Z=(x)/ ,( 8.26.4)/1.2=1.5 and(9.46.4)/1.2=2.5

we have:

P ( 8.2<X<9.4 )=P ( 1.5<Z<2.5 )=0.0606

The answer is "e"(0.06)

3)=6.4 and =1.2 we have:

P ( X<8.8 )=P ( X<8.86.4 )=P{ (X)/<(8.86.4)/1.2)}

Since (x)/=Z and( 8.86.4)/1.2=2 we have:

P (X<8.8)=P (Z<2)=0.9772

The answer is "d"(0.97)

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