Time spent using e-mail per session is normally distributed with a mean = 8 minutes and...

Time spent using e-mail per session is normally distributed with a mean = 8 minutes and standard deviation = 2 minutes. If a random sample of 36 sessions were selected, what percentage of sample means would be between 7.8 and 8.2?

a. .3546

b. .3333

c. .4514

d. .5021

Solutions

Expert Solution

Solution :

Given that,

mean = = 8

standard deviation = = 2

n = 36

= 8

= / n = 2 / 36 = 0.3333

P( 7.8 < < 8.2 ) = P((7.8 - 8 ) / 0.3333 <( - ) / < (8.2 - 8 ) / 0.3333 ))

= P( - 0.60 < Z < 0.60 )

= P(Z < 0.60 ) - P(Z < - 0.60 )

= 0.7257 - 0.2743

= 0.4514

Option (c) is correct

orchestra answered 1 year ago

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