Daniel Trumpe has computed the EOQ for a product he sells to be 400 units. However, due to recent events he has a cash flow prob
1 answer:
You might be interested in
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
Problems of normally distributed samples can be 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.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79
Using the z-distribution, it is found that the lower bound of the 99% confidence interval is given by:
d. 68.39%.
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 99% confidence level, hence, z is the value of Z that has a p-value of
, so the critical value is z = 2.575.
The sample size and estimate are given by:
Hence, the lower bound is given by:
Hence the lower bound is of 68.39%, which means that option D is correct.
More can be learned about the z-distribution at brainly.com/question/25890103
#SPJ1