Z = 1.555 should be used
If we seek an 88% confidence interval, that means we only want a 12% chance that our interval does not contain the true value.
Assuming a two-sided test, that means we want a 6% chance attributed to each tail of the Z-distribution.
the zα/2 value of z0.06.
This z value at α/2=0.06 is the coordinate of the Z-curve that has 6% of the distribution's area to its right, and thus 94% of the area to its left. We find this z-value by reverse-lookup in a z-table.
<h3>What is Z-distribution?</h3>
The standard normal distribution, also called the z-distribution, is a special normal distribution where the mean is 0 and the standard deviation is 1.
Any normal distribution can be standardized by converting its values into z-scores. Z-scores tell you how many standard deviations from the mean each value lies.
<h3>Why is z-score used?</h3>
The standard score (more commonly referred to as a z-score) is a very useful statistic because it
(a) allows us to calculate the probability of a score occurring within our normal distribution and
(b) enables us to compare two scores that are from different normal distributions.
To learn more about Z-distribution from the given link
brainly.com/question/17039068
#SPJ4
Answer:
Step-by-step explanation:
Demand = 19,500 units per year (D)
Ordering cost = $25 per order (O)
Holding cost = $4 per unit per year (C)
a)
=
=
= 493.71044 ≈ 494
b) Annual holding cost =
=
= 4 × 247
= 988
c) Annual ordering cost =
=
= 25 × 39.47
= 986.75
AOQ = 494
Annual holding cost = 988
Annual ordering cost = 986.75
Wheres the rest of the problem? or is this it?
Answer:
A) 1:4
Step-by-step explanation:
#We first calculate the campus' area:
#We then calculate the campus' map area in square ft:
Ratio is a comparison between two dimensions:
Hence, the ratio of the area in square miles of the campus to the area in square feet of the map is 1:4