Can someone help me plsss.
1 answer:
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Answer:
a) .033 b) .4678 c) 100
Step-by-step explanation:
Distribution of miraculin produced is normal. To answer question we need to find the z-score of 100g, 110g and 120g miraculin
General formula for z score is given by
z= where
- X is the miraculin produced (120,110,100)
- M is the mean miraculin produced for a partucilar tomato plant, which is 105.3
- s is the standard deviation of the sample, which is 8.0
a)
z-score for 120 g miraculin is =1.8375 then
P(x > 120) = P(z>1.8375) = 0.033
b)
z-score for 110 g miraculin is = 0.5875
z-score for 100 g miraculin is = -0,6625
P(100 < x < 110) = P(x>100) - P(x>110) =0.7462 - 0,2784 = 0.4678
c)
if P(x < a) = 0.25 then corresponding z score for a is -0.6745. From the above formula we can calculate a:
=-0.6625 which gives a= 100