Please show working. Weighing cereal. Chocolate-coated Sugar Fun Blast cereal is
ID: 3226698 • Letter: P
Question
Please show working.
Weighing cereal. Chocolate-coated Sugar Fun Blast cereal is filled into boxes by weight, with the weight approximately Normally distributed with an average of 16 ounces and a standard deviation of 0.2 ounces. a. What is the probability that the box you buy (chosen at random) will weigh less than 15.5 ounces? b. What is the probability that the box you buy will weigh between 16 and 16.25 ounces? c. What is the probability that the box you buy will weigh more than 17 ounces? d. What is the upper cut-off for the 90th percentile (bottom 90%) of weights? e. What is the range for the middle 60% of weights? f. What is the range for the upper 2.5%?Explanation / Answer
The weight here has a mean of 16 ounces and standard deviation of 0.2 ounces.
a) For X = 15.5
The z score would be:
z = (15.5 - 16) / 0.2 = -2.5
Therefore the required probability would be: P(Z < -2.5 )
Getting the above probability from the standard normal table we get:
P(Z < -2.5 ) = 0.0062
Therefore 0.0062 is the required probability here.
b) As 16 is the mean , the z score for mean would be 0.
Now for X = 16.25, the z score would be:
Z = (16.25 - 16)/0.2 = 1.25
Therefore the required probability would be:
P( 0 < Z < 1.25) = P( Z< 1.25) - P( Z< 0.5) = 0.8944 - 0.5 = 0.3944
Therefore 0.3944 is the required probability here.
c) For X = 17 ounces, the z value would be:
Z = (17-16)/0.2 = 5
Therefore the required probability here would be : P( Z > 5) = 0
Therefore approx. 0 is the required probability here.
d) Let the upper cutoff for the 90th percentile of weights be K.
Then P( X< K ) = 0.9
From standard normal tables we get:
P( Z< 1.282 ) = 0.9
Therefore we get:
(K - 16)/ 0.2 = 1.282
Therefore we get: K = 16 + 0.2*1.282 = 16.2564
Therefore the 90th percentile of weights here is 16.2564
e) Let the range for the middle 60% of the weights be (16 - K , 16 + K )
Then we are given that:
P( 16 - K < X< 16 + K ) = 0.6
Therefore due to symmetry we get:
P ( 16 < X < 16+ K ) = 0.3
P( X< 16 + K ) = 0.5 + 0.3= 0.8
From standard normal table we get:
P( Z< 0.8416 )= 0.8
Therefore we get:
(16 + K - 16) / 0.2 = 0.8416
K = 0.2*0.8416 = 0.16832
Therefore the middle 60% of the values lies in the range ( 16 - 0.16832, 16 + 0.16832 )
that is : ( 15.83168, 16.16832 )
This is the required range
f) Let the upper 2.5% values lie to the right of K. Then we have:
P( X > K ) = 0.025
From standard normal tables we get:
P( Z > 1.96 ) = 0.025
Therefore we get:
(K- 16) / 0.2 = 1.96
K = 16 + 0.2*1.96 = 16.392
Therefore the required range here is from 16.392 to infinity.