Question #16877

Unisys.com is one of the most frequented business to business web sites. according to a wall street journal article, business partners accessing Unisys.com spend an average of 65.7 minutes, possibly the longest average time per visit of any business to business website. assuming that the duration of a visit to this website is normally distributed with mean of 65.7 minutes and the standard deviation of 15 minutes. (A) what is the probability that the randomly selected visit will last between 60 and 90 minutes. (B) between what values, symmetrically distributed around the mean will90% of visit last?

Expert's answer

(a)

Z(60) = (60 - Mean)/SD = (60 – 65.7)/15 = -0.38

Z(90) = (90 – 65.7)/15 = 1.62,

Use z – table:

P(60 < time < 90) = p(-0.38 < z < 1.62) = p(0 < z < 0.38)+ p(0 < z < 1.62) = 0.148 + 0.4474 = 0.5954

(b)

Let these values be Mean + t and Mean – t.

So, p(-t < time - Mean < t) = 0.9 = p(z(-t) < z(Mean = 0) <z(t)) = 2*p(0 < z(Mean = 0) <z(t)) => z(t) = z(p(0 < z(Mean = 0) < z(t) = 0.9/2 = 0.45) = 1.645

z(t) = (t – 0)/SD => t = z(t)*SD + 0 = 1.645*15 ≈ 24.7

Mean – t = 41; Mean + t = 90.4

So, 90% of visit will last between 41 and 90.4 min

Z(60) = (60 - Mean)/SD = (60 – 65.7)/15 = -0.38

Z(90) = (90 – 65.7)/15 = 1.62,

Use z – table:

P(60 < time < 90) = p(-0.38 < z < 1.62) = p(0 < z < 0.38)+ p(0 < z < 1.62) = 0.148 + 0.4474 = 0.5954

(b)

Let these values be Mean + t and Mean – t.

So, p(-t < time - Mean < t) = 0.9 = p(z(-t) < z(Mean = 0) <z(t)) = 2*p(0 < z(Mean = 0) <z(t)) => z(t) = z(p(0 < z(Mean = 0) < z(t) = 0.9/2 = 0.45) = 1.645

z(t) = (t – 0)/SD => t = z(t)*SD + 0 = 1.645*15 ≈ 24.7

Mean – t = 41; Mean + t = 90.4

So, 90% of visit will last between 41 and 90.4 min

## Comments

Assignment Expert31.10.14, 17:09Dear julinda.You're welcome. We are glad to be helpful. If you really liked our service please press like-button beside answer field. Thank you!

julinda31.10.14, 14:13thanx

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