Question 1205250




given:

mean, {{{mu = 6.5}}} pounds

standard deviation, {{{sigma = 1.8}}}  pounds


a normal distribution formula:

{{{z=(x-mu)/sigma}}}


1. How many would you expect to weigh between {{{3 }}}and {{{7}}} pounds?

{{{P(3<=x<=7)}}}

={{{P((3-6.5)/1.8<=z<=(7-6.5)/1.8)}}}

={{{P(-1.9444444444444446<=z<=0.2777777777777778)}}}

={{{P(z<=0.2777777777777778) -P(z<-1.9444444444444446)}}}

={{{0.609409-0.0259209}}}

={{{0.5834881}}}

={{{58.35}}}% => answer

Consider a group of {{{1300}}} newborn babies, then {{{1300*0.5834881=758.53453}}}≈{{{758}}} babies expected to weigh between {{{3 }}}and {{{7}}} pounds


2. How many would you expect to weigh less than {{{6}}} pounds?


{{{P(x<6)=P(z>(6-6.5)/1.8)=P(z< -0.2777777777777778)=0.390591=39.06}}}%=> answer

{{{1300*0.390591}}}≈{{{507}}} babies  expected to weigh less than {{{6}}} pounds


3. How many would you expect to weigh more than {{{4}}} pounds?

{{{P(x>4)=P(z<(4-6.5)/1.8)=P(z>-1. 38889)=0.917567=91.76}}}% => answer

{{{1300*0.917567}}}≈{{{1192}}} babies expected to weigh more than {{{4}}} pounds


4. How many would you expect to weigh between {{{6.5}}} and {{{10}}} pounds?


{{{P(6.5<=x<=10)}}}

={{{P((6.5-6.5)/1.8<=z<=(10-6.5)/1.8)}}}

={{{P(0<=z<=1.9444444444444446)}}}

={{{P(z<=1.9444444444444446)-P(z>0)}}}

={{{0.974079-0.5}}}

={{{0.4741}}}

={{{47.41}}}% => answer

{{{1300*0.4741}}}≈{{{616}}} babies expected to weigh between {{{6.5}}} and {{{10}}} pounds