SOLUTION: suppose that the weiht of mangoes is normally distributed with mean 8 ounces and standard deviation is 1.5 ounces. a)what proportion of mangoes weigh more than 11.5 ounces b.wha

Question 1092748: suppose that the weiht of mangoes is normally distributed with mean 8 ounces and standard deviation is 1.5 ounces.
a)what proportion of mangoes weigh more than 11.5 ounces
b.what proportiion of mangoes weith less than 8.7 ounces.
c. what proportion of mangoes weith less than 5 ounces
d.what proportion of mangoes weith more than 4.9 ounces
e. what proportion of mangoes weith between 6.2 and 7 ounces
f. what proportion of mangoes weith between 10.3 and 14 ounces
g. what proportion of mangoes weith between 6.8and 8.9 ounces
h. find the 80th percentile of the distribution of x.
i. find the 5th percentile of the distribution of x.
j. find the interquartile range of the distribution of x.
show your working and formulas tx
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The basic formula is z=(x=mean)/std deviation.
a. z=(11.5-8)/1.5 so z=2.33. Want z > 2.33 From calculator or table, this probability is 0.0099.
b. z is less than .7/1.5 or 0.47. This is a probability of 0.3192.
c. z is -3/1.5 or z is less than -2, with probability of 0.0228
d. z is greater than -3.1/1.5=z is greater than -2.07 or probability 0.9808
e. this is a z between -1.8/1.5 and -1/1.5; this is -1.2 f. this is a z between 2.3/1.5 and 4 or probability of 0.0626
g. this is a z between -1.2/1.5 and +0.9/1.5 or between -0.8 and +0.6 or probability of 0.5139
h. 80th percentile is where z is +0.84, so that (x-mean)/1.5=0.84 or (x-mean=1.26), or x=mean+1.26, so that x=9.26 oz.
i. 5th percentile is z=-1.645, and x-mean = -2.47 so that x=8-2.47 or 5.53 oz.
j. interquartile is z.25 and z.75 which is -0.675 to +0.675 for z or +/- 1.01 oz from the mean or (6.99, 9.01). The +/- 1.01 is z* sd.

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