Average talk time between charges of a cell phone is advertised as 3 hours. Assu

Question

Average talk time between charges of a cell phone is advertised as 3 hours. Assume that talk time is normally distributed with a standard deviation of 0.6 hour. Use Table 1.

a. Find the probability that talk time between charges for a randomly selected cell phone is below 1.8 hours. (Round "z" value to 2 decimal places and final answer to 4 decimal places.) Probability

b. Find the probability that talk time between charges for a randomly selected cell phone is either more than 3.6 hours or below 1.4 hours. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

Probability

c. Twenty nine percent of the time, talk time between charges is below a particular value. What is this value? (Round "z" value to 2 decimal places and final answer to 3 decimal places.) Value

Explanation / Answer

Here mean talk time between charges of a cell phone = 3 hours

Sandard deviation of talks = 0.6 hours

(a) Here we have to find if X is the talk time between charges for a randomly selected cell phone

Pr(X < 1.8) = NORM (X < 1.8 ; 3 ; 0.6)

Z = (1.8 - 3.0)/ 0.6 = -2.00

Pr(X < 1.8) = Pr(Z < -2.00) = 0.0228

(b) Now we have to find

Pr(X > 3.6 Hours and X < 1.4 hours) = 1 - Pr(X < 3.6 hours ) + Pr(X < 1.4 Hours) =1 - (Z2 ) + (Z1)

Z2 = (3.6 - 3)/0.6 = 1

Z1 = (1.4 - 3)/ 0.6 = -2.67

Pr(X > 3.6 Hours and X < 1.4 hours) =1 - (Z2 ) + (Z1) = 1 - 0.8413 + 0.0038 = 0.1625

(c) Here Pr(X < K) = 0.29

Let say that particulat value is K

so Pr(X < K ; 3 ; 0.6) = 0.29

Z value from Z table for the given p - value

Z = -0.5534

(K - 3)/0.6 = -0.5534

K = 3 - 0.6 * 0.5534 = 2.668 hours

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