Question

A vegetable distributor knows that during the month of August ,the weights of tomatoes are normally distributed with a mean of 0.61 lb and a standard deviation of 0.15 lb. How many can be expected to weigh between 0.31 to 0.91 in a shipment of 4500 tomatoes.

Select one:
a. 4000
b. 4275
c. 4100
d. 4215

Answer 1

Answer:

first find normal variable using

Z = (X - mean)/standard deviation

so corresponding to x = 0.31

Z1 = (0.31 - 0.61)/0.15 = - 2

and corresponding to x = 0.91

Z2 = (0.91 - 0.61)/0.15 = 2

Required area =

$p( - 2 \leqslant z \leqslant 2) = 2p(0 \leqslant z \leqslant 2) = 2 \times 0.4772 = 0.9542$

so weight of 4500 tomatoes = 4500 x 0.9542

so approximately 4275

option b is correct

Answer 2

Given:

A vegetable distributor knows that during August, the weights of tomatoes are normally distributed with a mean of $0.61 lb$ and a standard deviation of $0.15 lb$.

To Find:

How many can be expected to weigh between $0.31 to 0.91$ in a shipment of $4500$ tomatoes

Step-by-step explanation:

There are $5863.5$ towards can be expected to weigh more than $0.3116$

$0.61-0.15*2=0.31$

$0.61+0.15*2=0.91$

$=4500*95.45$

$=4275$

Answer:

Therefore, The weight $0.31$to $0.91$ in a shipment of $4500$ tomatoes in the $4275$.