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To find the z-score for a weight of 196 oz., use
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
A table for the cumulative distribution function for the normal distribution (see picture) gives the area 0.9772 BELOW the z-score z = 2. Carl is wondering about the percentage of boxes with weights ABOVE z = 2. The total area under the normal curve is 1, so subtract .9772 from 1.0000.
1.0000 - .9772 = 0.0228, so about 2.3% of the boxes will weigh more than 196 oz.
Answer:
Step-by-step explanation:
Hello!
You have the information for two variables
X₁: Number of consumer purchases in France that were made with cash, in a sample of 120.
n₁= 120 consumer purchases
x₁= 48 cash purchases
p'₁= 48/120= 0.4
X₂: Number of consumer purchases in the US that were made with cash, in a sample of 55.
n₂= 55 consumer purchases
x₂= 24 cash purchases
p'₂= 24/55= 0.4364
You need to construct a 90% CI for the difference of proportions p₁-p₂
Using the central limit theorem you can approximate the distribution of both sample proportions p'₁ and p'₂ to normal, so the statistic to use to estimate the difference of proportions is an approximate standard normal:
[(p'₁-p'₂) ± *
]
[(0.4-0.4364)±1.648 * ]
[-0.1689;0.0961]
The interval has a negative bond, it is ok, keep in mind that even tough proportions take values between 0 and 1, in this case, the confidence interval estimates the difference between the two proportions. It is valid for one of the bonds or the two bonds of the CI for the difference between population proportions to be negative.
I hope this helps!