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 8.5 pounds and a variance of 1.21 pounds2. A bag of California oranges is randomly selected in the grocery store.

(Round all probability answers to four decimal places.)

a. What is the probability that a randomly selected California orange bag purchased by a customer weighs more than 6 pounds?

b. What is the probability that a randomly selected California orange bag purchased by a customer weighs between 6.9 and 8 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. 12% of the time, a customer will buy a bag of California oranges that weighs more than a specific weight. Find that weight.

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 8.5
standard Deviation ( sd )= 1.1(since variance 1.21 was given)
a.
P(X > 6) = (6-8.5)/1.1
= -2.5/1.1 = -2.2727
= P ( Z >-2.2727) From Standard Normal Table
= 0.9885

b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 6.9) = (6.9-8.5)/1.1
= -1.6/1.1 = -1.4545
= P ( Z <-1.4545) From Standard Normal Table
= 0.0729
P(X < 8) = (8-8.5)/1.1
= -0.5/1.1 = -0.4545
= P ( Z <-0.4545) From Standard Normal Table
= 0.3247
P(6.9 < X < 8) = 0.3247-0.0729 = 0.2518

c.
P(X = 5) = (5-8.5)/1.1
= -3.5/1.1= -3.1818
= P ( Z =3.1818) From Standard Normal Table
= 0.0007

d.
P ( Z > x ) = 0.12
Value of z to the cumulative probability of 0.12 from normal table is 1.174987
P( x-u / (s.d) > x - 8.5/1.1) = 0.12
That is, ( x - 8.5/1.1) = 1.174987
--> x = 1.174987 * 1.1+8.5 = 9.79

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