The ORAC (oxygen radical absorbing capacity) in a ripe highbush blueberry fruit

Question

The ORAC (oxygen radical absorbing capacity) in a ripe highbush blueberry fruit is known to be normally distributed with mean 423 MTE (micromole trolox equivalents) with a population standard deviation of 51.7 MTE. You should not round intermediate calculations. Once you find your Z-scores, then you can round the Z-scores to two decimal places before finding a probability. What is the probability that the fruit on a highbush blueberry bush would have an ORAC between 423 and 500 MTE (4 decimal places)? What is the probability that the fruit on a highbush blueberry bush would have an ORAC of less than 300 MTE (4 decimal places)? If you randomly sampled 400 highbush blueberries, approximately how many of these berries would have an ORAC between 450 and 470 MTE? (round to the nearest whole number). What is the ORAC of the blueberries that bound the middle 80% of the ORAC of all the blueberries on a highbush plant (round final answers to 1 decimal Place)? Smallest value: MTE Largest value: MTE

Explanation / Answer

1) mean =423

std deviation=51.7

from normal distribution zscore =(X-mean)/std deviation

a)hence P(423<X<500)=P((423-423)/51.7<Z<(500-423)/51.7)=P(0<Z<1.4894)=0.9318-0.5=0.4318

b)P(X<300)=P(Z<(300-423)/51.7)=P(Z<-2.379)=0.0087

c)P(450<X<470)=P(0.9091<Z<0.5222)= 0.8183-0.6993=0.1191

hence number of berries=np=400*0.1191=48(rounding it up)

d)for middle 80%, z=+/-1.2816

hence interval =mean +/- z*std deviation =356.7 ; 489.3

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.