Lennie Davis
2021-12-12
Lynne Trussell
Beginner2021-12-13Added 32 answers
Step 1
(a) If the diameter of the straw is less than 3 mm, it can fit in the hole. The standard deviation is 0.25 mm, and the mean is 2.6 mm.
The likelihood that it will fit in the hole in a beverage carton is,
The probability of z less than 1.60 can be obtained using the excel formula “=NORM.S.DIST(1.60,TRUE)”. The probability value is 0.9542.
The required probability is,
Therefore, the probability that it fits into the hole in a drinks carton is 0.9542.
Step 2
(b) Let x be the number of straws fit into the holes in drinks cartons which follows binomial with sample size 500 and probability of success 0.9542.
The mean is,
The standard deviation is,
Using a binomial model with a sample size of 500 and a success probability, let x be the number of straws that fit into the holes in beverage cartons,
[Continuity correction]
The probability of z less than 0.5134 can be obtained using the excel formula “=NORM.S.DIST(0.5134,TRUE)”. The probability value is 0.6962.
The required probability is,
Hence, the probability that at least 480 straws fit into the holes in drinks cartons is 0.3038.
Fasaniu
Beginner2021-12-14Added 46 answers
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