The Hu family goes out for lunch, and the price of the meal is $52. The sales tax on the meal is 6%, and the family also leaves
2 answers:
You might be interested in
Answer:
The probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
Step-by-step explanation:
Given:
Weight of a given sample (x) = 2.33 oz
Mean weight (μ) = 1.75 oz
Standard deviation (σ) = 0.22 oz
The distribution is normal distribution.
So, first, we will find the z-score of the distribution using the formula:
Plug in the values and solve for 'z'. This gives,
So, the z-score of the distribution is 2.64.
Now, we need the probability .
From the normal distribution table for z-score equal to 2.64, the value of the probability is 0.9959. This is the area to the left of the curve or less than z-score value.
But, we need area more than the z-score value. So, the area is:
Therefore, the probability of the stick's weight being 2.33 oz or greater is 0.0041 or 0.41%.
Answer:
We need an SRS of scores of at least 153.
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
How large an SRS of scores must you choose?
This is at least n, in which n is found when . So
Rounding to the next whole number, 153
We need an SRS of scores of at least 153.