in a grocery store, butter is sold in “sticks” that are shaped like little bricks. The weights of these sticks are like draws at random with replacement from a population with average 4 ounces and SD 0.2 ounces. The grocery store receives the butter in boxes; each box consists of 100 sticks.
1.The chance that the average weight of the sticks in one box is less than 3.999 ounces is closest to (pick the best of the available options, even if you can calculate the chance more accurately)
32%
36%
40%
44%
48%
50%
52%
2.The grocery store has received 6 boxes of butter. There is about ___________ chance that in at least one of the boxes, the average weight of sticks is less than 3.999 ounces. Fill in the blank with the best of the following options.
82%
86%
90%
94%
98%
Pls help
1.50%
2.82%
1)average=4; SD=0.2;sample =100
the expected total = 100*4=400
SE of total= sqrt(100)*0.2=2
(100*3.999-400)/2=-0.05
using applet:-5...0.05
approximately 48%
1)average=4; SD=0.2;sample =100
the expected total = 100*4=400
SE of total= sqrt(100)*0.2=2
(100*3.999-400)/2=-0.05
using applet:-5...-0.05
approximately 48%
2) binomial:
n=6 p=0.48; x=6
using applet:1.22306%
1-1.22306%=98%
its right dean thanks
1. To calculate the chance that the average weight of the sticks in one box is less than 3.999 ounces, we will use the z-score formula.
The z-score formula is defined as:
z = (x - μ) / σ
Where:
x = the value we want to find the probability for (3.999 ounces)
μ = the population mean (4 ounces)
σ = the standard deviation of the population (0.2 ounces)
Now, let's calculate the z-score for 3.999 ounces:
z = (3.999 - 4) / 0.2
z = -0.001 / 0.2
z ≈ -0.005
Next, we need to find the probability corresponding to this z-score using a standard normal distribution table (also known as a z-table). We want to find the probability to the left of -0.005.
Looking up the z-score -0.005 in the standard normal distribution table, we find that the probability is approximately 0.499.
Now, since the weights of the sticks are drawn at random with replacement from a population, the average weight of the sticks in each box follows a normal distribution as well. Therefore, the chance that the average weight of the sticks in one box is less than 3.999 ounces is equal to the probability we calculated, which is approximately 0.499 or 49.9%.
Therefore, the closest option from the given choices is 50%.
2. To calculate the chance that in at least one of the 6 boxes, the average weight of sticks is less than 3.999 ounces, we need to calculate the complementary probability.
Calculating the probability that the average weight of sticks in one box is equal to or greater than 3.999 ounces is (1 - 0.499) = 0.501.
Since the boxes are independent, the probability that the average weight of sticks in all 6 boxes is equal to or greater than 3.999 ounces is 0.501^6 ≈ 0.032.
Therefore, the chance that in at least one of the 6 boxes, the average weight of sticks is less than 3.999 ounces is (1 - 0.032) = 0.968.
Rounding to the closest option from the given choices, we get 98%.
So, the answer for the second question is 98%.