Question

A potato sorting machine produces 5 lb bags of potatoes on average, but the standard deviation is 1/2 lb. Assuming that the weight of the bags is normally distributed, explain how you would find the weight range that 95% of the bags fall within.

Answer 1

Basically we have to calculate P X < 0.95. By using the normal distribution formula for any random variable we can get our answer.

SO first from z table we will have to check the corresponding z value from the table. After we get that value we can do X - mean / the std dev = that particular value that i am naming A.

SO X - 5 / 0.5 = A. Now we can solve X for our answer.

Answer 2

Answer:

The range that 95% of the bags fall within is (4.02,5.98)            

Step-by-step explanation:

Given : A potato sorting machine produces 5 lb bags of potatoes on average, but the standard deviation is 1/2 lb. Assuming that the weight of the bags is normally distributed

To find : How you would find the weight range that 95% of the bags fall within?

Solution :

We have given,

Mean is $\mu=5$

Standard deviation is $\sigma=\farc{1}{2}=0.5$

The critical value for a 95% confidence interval is 1.96.

Let x=1.96

The formula to find the range or confidence interval is

$\mu-x\sigma\leq CI\leq \mu+x\sigma$

Substitute the value in the formula,

$5-1.96\times 0.5\leq CI\leq5+1.96\times 0.5$

$5-0.98\leq CI\leq5+0.98$

$4.02 \leq CI \leq 5.98$

Therefore, The range that 95% of the bags fall within is (4.02,5.98)