A grocery store purchases bags of oranges from California to sell in their store

Question

A grocery store purchases bags of oranges from California to sell in their store. The weight of a bag of California oranges is normally distributed with a mean of 7.7 pounds and a variance of 1.21 pounds2. A bag of California oranges is randomly selected in the grocery store.

a. What is the probability that a randomly selected California orange bag purchased by a customer weighs more than 8 pounds?

b. What is the probability that a randomly selected California orange bag purchased by a customer weighs between 6.1 and 7.2 pounds?

c. What is the probability that a randomly selected California orange bag purchased by a customer weighs exactly 5 pounds?

(Round weight to two decimal places)

d. 20% of the time, a customer will buy a bag of California oranges that weighs more than a specific weight. Find that weight.

Explanation / Answer

mean = 7.7 , variance = 1.21 , std.devition = sqrt(variance) = 1.1

a) P(X > 8)

z = ( x - mean ) / s

= ( 8 - 7.7 ) / 1.1

= 0.27

Now, we need to find P(Z > 0.27)

P(X > 8) = P(Z > 0.27) = 0.3925

b)

P( 6.1 < X < 7.2)

P(X< 6.1)

z = ( x - mean ) / s

= ( 6.1 - 7.7 ) / 1.1

= -1.45

P(X < 7.2)

z = ( x - mean ) / s

= ( 7.2 - 7.7 ) / 1.1

= - 0.45

P(6.1 < X < 7.2) = P( - 1.45 < z < -0.45) = 0.2518

c)

P(4.95 < X <5.05)

P(X < 4.95)

z = ( x - mean ) / s

= ( 4.95 - 7.7 ) / 1.1

= - 2.5

P(X < 5.05)

z = ( x - mean ) / s

= ( 5.05 - 7.7 ) / 1.1

= - 2.41

P(4.95 < X <5.05) = P( - 2.5 < z < -2.41) = 0.0018

d) z value for 20% = 0.8416

X bar = mean + z * s

= 7.7 + 0.8416 * 1.1

= 8.626

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