TakeTest 3icholas Noble . Mcmoft Edge rowst Edge STA 220 STA 220 Sulivan Se-7521

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TakeTest 3icholas Noble . Mcmoft Edge rowst Edge STA 220 STA 220 Sulivan Se-7521 242 TA 220 Sullvan se Test: Test Four (chapters 7,8,9) This Question: 1 pt The number of chocolate chips in an 18 ounce bag of choceate hip coiessaprmaliaty nmaly (a) What is the probability that a randomly selecled bag contanrs baheen 000 and 1400 by What is the probablty that a randomly selicted bag cantaine ha 1125 checolale chipa? cookies is apprmalaty rmaly Sistrbuted with mean 1252 and standard deviations 129 chips (c) what preposen of bags contains morean t200 choelate aus? (d) What is the percentle rank of a bag that contaies 1025 chacolale chips? chacolate chipn hp?e 1252 and Cick the icom to vidw a table of areas under thie nomel orve Round to fo Bat randon Round so four decimal places as needed a) The probability hat . .andomly solected beg contain" biteren oto and 400 chocouchp th) The prabatny that a randomly selected bag eentans tower than 112S cnocelate chips a ound to four dacimal places as needed ) as needed ower Round to fecr decimai places as neaded ) MA bag tha, contains 1126 chocolate chips is in tho L jmpencemia Round 10 he newest rteger as needed

Explanation / Answer

Solution:

a) = 1252
= 129
standardize x to z = (x - ) /
P( 1000 < x < 1400) = P[( 1000 - 1252) / 129 < Z < ( 1400 - 1252) / 129]
P( -1.9535 < Z < 1.1473) = P( z < 1.1473) - P( z < -1.9535) = 0.8749 - 0.0256 = 0.8493
(From Normal probability table)

b)
= 1252
= 129
standardize x to z = (x - ) /
P(x < 1125) = P( z < (1125-1252) / 129)
= P(z < -0.9845) = 0.1635
(From Normal probability table)

c)
= 1252
= 129
standardize x to z = (x - ) /
P(x > 1225) = P( z > (1225-1252) / 129)
= P(z > -0.2093) = 0.5832
(From Normal probability table)

d)
= 1252
= 129
standardize x to z = (x - ) /
P(x < 1025) = P( z < (1025-1252) / 129)
= P(z < -1.7597) = 0.0392
(From Normal probability table)
Percentile rank is 3.92 or 4

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