(Due 7/24) The number of chocolate chips in an 18-ounce bag is approximately nor
ID: 2949656 • Letter: #
Question
(Due 7/24) The number of chocolate chips in an 18-ounce bag is approximately normally distributed with a mean of 1262 chips and standard deviation 118 chips according to a study, a) What is the probability that a randomly selected 18-ounce bag contains between 1. 1000 and 1400 chips? b) What is the probability that a randomly selected 18-ounce bag contains fewer than 990 chips? c) What is the probability that a randomly selected 18-ounce bag contains more than 1200 chips? d) What is the percentile rank of an 18-ounce bag that contains 1050 chips? el What is the interquartile range of the number of chips? f) Determine the chips of the bottom 5%. g) Determine the chips of the top 10%.Explanation / Answer
1.
We are given that Mean = 1262 , Standard deviation = S = 118
a) We need to find P( 1000 < X < 1400 )
= P( X < 1400 ) - P( X < 1000 )
= P( (X - mean) / S < (1400 - Mean) / S ) - P( (X - mean) / S < (1000 - Mean) / S )
= P( Z < (1400 - 1262) / 118 ) - P( Z < (1000 - 1262) / 118 )
= P( Z < 1.17) - P ( Z < -2.22)
= 0.8658
b) We need to find , P( X < 990 )
= P( (X - mean) / S < (990 - Mean) / S )
= P( Z < (990 - 1262 ) /118)
= P(Z< -2.305)
= 0.0106
c) We need to find, P( X > 1200)
= 1- P( X < 1200 )
= 1 - P( (X - mean) / S < (1200 - Mean) / S )
= 1 - P( Z < ( 1200 - 1262 ) / 118 )
= 1 - P(Z < -0.5254)
= 0.7
d) We have P( X < 1050) = Y
We need to find Y
P( X < 1050) = Y
=> P( (X - mean) / S < (1050- Mean) / S ) = Y
=> P ( Z < -1.7966 ) = Y
=> Y = 0.0362
Percentile rank for an 18-ounce bag that contains 1050 chips = 0.0362 = 3.62 %
e) Interquartile range = Q3 - Q1
Where , Q3 = third quartile
Q1 = First quartile
P( X < Q1 ) = 0.25
P( (X - mean) / S < (Q1- Mean) / S ) = 0.25
P( Z < (Q1- 1262) / 118 ) = 0.25
( Q1 - 1262 ) / 118 = -0.6745
=> Q1 = 1182.41
P( X < Q3 ) = 0.75
P( (X - mean) / S < (Q3- Mean) / S ) = 0.75
P( Z < (Q3- 1262) / 118 ) = 0.75
( Q3 - 1262 ) / 118 = 0.6745
=> Q3 = 1341.59
Interquartile range (IQR) = Q3 - Q1 = 1341.59 - 1182.41 = 159.181
f) Let us denote the chips below 'B' value to be bottom 5%
We have P( X < B ) = 0.05
We need to find 'B'
=> P( (X - mean) / S < (B- Mean) / S ) = 0.05
=> P( Z < (B- 1262) / 118 ) = 0.05
=> (B- 1262) / 118 = -1.645
=> B = 1067.89
Any number of chips below 1067.89 falls in bottom 5%
g)
Let us denote the chips above 'T' value to be top 10%
We have P( X > T ) = 0.10
1- P( X < T )= 0.10
P( X < T ) = 0.90
We need to find 'T'
=> P( (X - mean) / S < (T- Mean) / S ) = 0.90
=> P( Z < (T- 1262) / 118 ) = 0.90
=> (T- 1262) / 118 = 1.28155
=> T = 1413.22
Any number of chips above 1413.22 falls in top 10%
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