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olga2289 [7]
3 years ago
14

The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard de

viation of 2.2 minutes. Find the probability that a call a. lasts between 5 and 10 minutes. b. lasts more than 7 minutes. c. lasts less than 4 minutes.
Mathematics
1 answer:
qaws [65]3 years ago
8 0

Answer: a. 0.6759    b. 0.3752    c. 0.1480

Step-by-step explanation:

Given : The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes

i.e. \mu = 6.3 minutes

\sigma=2.2 minutes

Let x be the long-distance call length.

a. The probability that a call lasts between 5 and 10 minutes will be :-

P(5

b. The probability that a call lasts more than 7 minutes. :

P(X>7)=P(\dfrac{X-\mu}{\sigma}>\dfrac{7-6.3}{2.2})\\\\=P(Z>0.318)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z

c. The probability that a call lasts more than 4 minutes. :

P(X

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Then isolate 'x' by doing reverse operations (on both sides) for every number that is on the same side as 'x'. Reverse operations are opposites (addition and subtraction, multiplication and division) and "cancel out" a number, which "moves" it to the other side.

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Check your answer:

Substitute 'x' for 41 in the original equation. Then, make sure that the equation is still balanced. Keep in mind BEDMAS order (brackets, exponents, division, multiplication, addition, subtraction).

7(x - 5) = 252

7(41 - 5) = 252        Simplify inside the brackets first

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LS = RS

Since left side (LS) equals (=) right side (RS), the equation is still balanced and we got the correct value for 'x'.

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