Answer:
Weights of at least 340.1 are in the highest 20%.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

a. Highest 20 percent
At least X
100-20 = 80
So X is the 80th percentile, which is X when Z has a pvalue of 0.8. So X when Z = 0.842.




Weights of at least 340.1 are in the highest 20%.
Answer:4.6h-2.9d-16
Step-by-step explanation:
Answer:
E) we will use t- distribution because is un-known,n<30
the confidence interval is (0.0338,0.0392)
Step-by-step explanation:
<u>Step:-1</u>
Given sample size is n = 23<30 mortgage institutions
The mean interest rate 'x' = 0.0365
The standard deviation 'S' = 0.0046
the degree of freedom = n-1 = 23-1=22
99% of confidence intervals
(from tabulated value).





using calculator

Confidence interval is


the mean value is lies between in this confidence interval
(0.0338,0.0392).
<u>Answer:-</u>
<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>
2(3r+7)−(2+r)
Use distributive property:
6r + 14 - 2 - r
Simplify by combining like terms:
5r +12
B = 16
explanation:
b-10 > 5
16-10 > 5
6 > 5
The other answer choices are less than or equal to 5