Question

Doggie Nuggets Inc.​ (DNI) sells large bags of dog food to warehouse clubs. DNI uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed with a mean of 46 kilograms and a standard deviation of 1.68 kilograms. Complete parts a through d below.

What is the probability that a filled bag will weigh less than 45.6 ​kilograms?

The probability is _______________.

​(Round to four decimal places as​ needed.)

b. What is the probability that a randomly sampled filled bag will weigh between 44.6 and 48 ​kilograms?

The probability is ____________________.

​(Round to four decimal places as​ needed.)

c. What is the minimum weight a bag of dog food could be and remain in the top 10​% of all bags​ filled?

The minimum weight is ________________ kilograms.

​(Round to three decimal places as​ needed.)

d. DNI is unable to adjust the mean of the filling process.​ However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that no more than 8​% of all filled bags weigh more than 51 ​kilograms?

The standard deviation would need to be _______________ kilograms.

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 46

standard deviation = = 1.68

a)

P(x < 45.6) = P((x - ) / < (45.6 - 46) / 1.68)

= P(z < -0.24)

= 0.4052

Probability = 0.4052

b)

P(44.6 < x < 48) = P((44.6 - 46)/ 1.68) < (x - ) /  < (48 - 46) / 1.68) )

= P(-2.024 < z < 1.19)

= P(z < 1.19) - P(z < -2.024)

= 0.883 - 0.0215

= 0.8615

Probability = 0.8615

(c)

Using standard normal table,

P(Z > z) = 10%

1 - P(Z < z) = 0.10

P(Z < z) = 1 - 0.10

P(Z < 1.28) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 1.68 + 46 = 48.15

The minimum weight is 48.15 kilograms .

(d)

x = 51

z = 1.405

x = z * +

= (x - ) / z = (51 - 46) / -1.405 = 3.559 kilograms