The actual weights of bag of pet food are normally distributed.The mean of the weights is 50.0 lb,with a standard deviation of 0.2 lb.
A)About what percent of bags of pet food weigh less than 49.8 lb?
The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 170 and a variance of 4. The material is considered defective if the breaking strength is less than 166 pounds. What is the probability that a single, randomly selected piece of material will be defective?
Well, I would say that the percentage of bags of pet food weighing less than 49.8 lb is about the same as the chances of finding a unicorn riding a unicycle on a rainbow. In other words, it's pretty rare!
To find the percentage of bags of pet food that weigh less than 49.8 lb, we can use the normal distribution and the given mean and standard deviation.
First, we need to calculate the z-score, which measures how many standard deviations an observation is from the mean. The formula for the z-score is:
z = (x - μ) / σ
where x is the value we want to calculate the probability for, μ is the mean, and σ is the standard deviation.
In this case, x = 49.8 lb, μ = 50.0 lb, and σ = 0.2 lb.
Let's calculate the z-score:
z = (49.8 - 50.0) / 0.2
z = -0.2 / 0.2
z = -1
Now, we need to find the area under the normal distribution curve to the left of the z-score. We can use a standard normal distribution table or a calculator to find this area.
Looking up -1 in a standard normal distribution table, we find that the area to the left of -1 is approximately 0.1587.
So, about 15.87% of bags of pet food weigh less than 49.8 lb.
To find the percentage of bags of pet food that weigh less than 49.8 lb, we need to calculate the z-score and then use a standard normal distribution table.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value we are interested in (49.8 lb), μ is the mean (50.0 lb), and σ is the standard deviation (0.2 lb).
z = (49.8 - 50.0) / 0.2 = -0.2 / 0.2 = -1
Next, we use the standard normal distribution table (also known as the z-table) to find the corresponding percentile for z = -1. A z-table provides the area under the standard normal curve to the left of a given z-score.
Looking up the z-score of -1 in the z-table, we find that the area to the left of -1 is approximately 0.1587 or 15.87%.
Therefore, about 15.87% of bags of pet food weigh less than 49.8 lb.
z=(X-μ)/σ
=(49.8-50)/0.2
=-1
Look up the probability table normal distributions for z=-1, which should give around 16%.