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Statistics 1 Question
Could someone solve this Statistics Question for me:
A market sells potatoes weighing more than 80 grams packed seperately packaged. Potatoes weighing between 80 and L grams are labelled as 'large', while potaotoes over L grams are labelled as 'extra large'.
Given that a randomly chosen potato is twice as likely to be 'large' as 'extra large', calculate the value of L.
Given data: Mean: 65 grams, Standard Deviation: 15 grams.
A market sells potatoes weighing more than 80 grams packed seperately packaged. Potatoes weighing between 80 and L grams are labelled as 'large', while potaotoes over L grams are labelled as 'extra large'.
Given that a randomly chosen potato is twice as likely to be 'large' as 'extra large', calculate the value of L.
Given data: Mean: 65 grams, Standard Deviation: 15 grams.
aiman
Could someone solve this Statistics Question for me:
A market sells potatoes weighing more than 80 grams packed seperately packaged. Potatoes weighing between 80 and L grams are labelled as 'large', while potaotoes over L grams are labelled as 'extra large'.
Given that a randomly chosen potato is twice as likely to be 'large' as 'extra large', calculate the value of L.
Given data: Mean: 65 grams, Standard Deviation: 15 grams.
A market sells potatoes weighing more than 80 grams packed seperately packaged. Potatoes weighing between 80 and L grams are labelled as 'large', while potaotoes over L grams are labelled as 'extra large'.
Given that a randomly chosen potato is twice as likely to be 'large' as 'extra large', calculate the value of L.
Given data: Mean: 65 grams, Standard Deviation: 15 grams.
80<X<L = large
X>L = extra large
P(X>80)
= P(Z>[80-65]/15)
= P(Z>1)
= 0.1587
Since a randomly chosen potato is twice as likely to be "large" as "extra large"
P(X>L) = 1/3 * (0.1587)
P(Z>[L-65]/15) = 0.0529
P(Z<[L-65]/15) = 0.9471
(L-65)/15 = 1.618
L = 89.3 (3 s.f.)
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