Question

A grocery store purchases bags of oranges from California to sell in their store. The weignt of a bag of Calomia cranges is nomally distributed with a mean of 7.6 pouncs and a variance of 121 pounds2 A bag of Califomia oranges is randomly selected in the grocery stone or Calremia Round all probabllty answers to four decimail placas.) what is the prabahiity that rannomly selerted Caltamaa range bag purchased by 。usic er w hs mre than 9 pounds b What is the protability that a rancomly selected Caliomia oange bag purchased by a custmer weighs belween G.9 and pounds? C. Wat is the prabatility Ihal a rardomty selecled Caifia orange tag purchaset by a customer weigts exaclly 5 paurds? Round weight o twe decimal places) d. 12% or the time, a customer All buy a bag orcallfcrnia oranges that weighs more than a specinc wegnt Find that weight welghs

Answer 1

ANSWER:-

a).We have o calculate the P(X>9):-

$P(X\leq 9)=P(Z\leq \frac{x-\mu }{s})$

Where,

X=9

$\mu =7.6$

$s^{2}=1.21$

$s=\sqrt{1.21}$

$s=1.1$

$P(Z\leq \frac{1.4}{1.1})$

$P(Z\leq 1.27)$

$P(Z> 9)=1-P(Z\leq 1.27)$

From Z table

$P(Z\leq 1.27)=0.8980$

$P(Z> 9)=1-0.8980$

$P(Z>9)=0.102$

Therefore the probability that a randomly selected california orange bag purchased by a customer weights more than 9 pounds is 0.102

b).We have o calculate the $P(6.9<X<8)$:-

$P(6.9<X<8)=P(Z<8)-P(Z<6.9)$

$P(X<8)=P(Z<\frac{x-\mu }{s})$

$P(X<8)=P(Z<\frac{8-7.6}{1.1})$

$P(X<8)=P(Z<\frac{0.4 }{1.1})$

From z table $P(Z<0.36)=0.6406$

$P(X<6.9)=P(Z<\frac{6.9-7.6}{1.1})$

$P(X<6.9)=P(Z<\frac{-0.7}{1.1})$

$P(X<6.9)=P(Z<-0.64)$

From z table $P(Z<-0.64)=0.2611$

$P(6.9<X<8)=0.6406-0.2611$

$P(6.9<X<8)=0.3795$

d). Here, we have to calculate the 12% of the time weights more than a specific weight:-

100%-12%=0.88

$P(X<\frac{x-\mu }{s})=0.88$

Where,

$\mu =7.6$

$s=1.1$

Then we get weight 9.068

NOTE:-

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