Suppose the time to transmit an e-mail message is normally distributed with a me

Question

Suppose the time to transmit an e-mail message is normally distributed with a mean of 0.75 seconds and a standard deviation of 0.2 seconds. Use 4DP in all your answers unless specifized. a. What is the probability that an e-mail message will require (1) more than one second to transmit? ______ (2) between 0.80 and 0.90 second to transmit? ______ b. What is the 70th percentile (30% of all the values exceed the X which is to be found) of the transmission time? ________ c. 90% of the messages will take at least how many seconds to be transmitted (find x such that P(X> =x)=0.9)?_______ Suppose that a sample of 20 e-mail message is slec d. What is the probability the average time for an e-mail message to be transmitted is (1) more than 0.8 second to transmit? _______ (2) between 0.80 and 0.90 second to transmit? _______ e. 95% of the average transmission times will be between what two values (assume symmetric about mean)? _______ and _______

Explanation / Answer

a 1)P(X>1)=1-P(X<1)=1-P(Z<(1-0.75)/0.2)=1-P(Z<1.25)=1-0.8944=0.1056

2)P(0.8<X<0.9)=P(0.25<X<0.75)=0.7734-0.5987=0.1747

b)for 70th percentile ; z=0.5244

hence X=0.75+0.2*0.5244 =0.85 seconds

c) for low 10%^,z =1.28155

hence X=0.75+0.2*1.2815 =0.49 seconds

d) for sample of 20, std error =std deviation/(n)1/2 =0.0447

1) P(X>0.8)=1-P(X<0.8)=1-P(Z<1.118)=1-0.8682=0.1318

2)P(0.8<X<0.9)=P(1.118<X<3.354)=0.9996-0.8682=0.1314

e) for 95% CI, z=1.96

hence confidence interval =0.66 ; 0.84

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