Answer: a) (176.76,172.24), b) 0.976.
Step-by-step explanation:
Since we have given that
Mean height = 174.5 cm
Standard deviation = 6.9 cm
n = 50
we need to find the 98% confidence interval.
So, z = 2.326
(a) Construct a 98% confidence interval for the mean height of all college students.
(b) What can we assert with 98% confidence about the possible size of our error if we estimate the mean height of all college students to be 174.5 centime- ters?
Error would be
Hence, a) (176.76,172.24), b) 0.976.
Answer:
C
Step-by-step explanation:
Mark as brainliest plz
Answer:
2
Step-by-step explanation:
3 plus 3 is 6 then divided by 3 is 2
Answer:
224 cubes
Step-by-step explanation:
The volume of the cubes is 1/64. Because 1/4*1/4*1/4=1/64.
The volume of the prims is 14/4 because 7/4*2*1=14/4.
Take the 14/4 and divide it by 1/64.
Which equals 896/4 = 224
Hope that helped.
Answer:
a) 0.4121
b) $588
Step-by-step explanation:
Mean μ = $633
Standard deviation σ = $45.
Required:
a. If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?
We solve using z score formula
= z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
For x = $646
z = 646 - 633/45
z = 0.22222
Probability value from Z-Table:
P(x<646) = 0.58793
P(x>646) = 1 - P(x<646) = 0.41207
≈ 0.4121
b. How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16? (Round your answer to the nearest dollar.)
Converting 0.16 to percentage = 0.16 × 100% = 16%
The z score of 16%
= -0.994
We are to find x
Using z score formula
z = (x-μ)/σ
-0.994 = x - 633/45
Cross Multiply
-0.994 × 45 = x - 633
-44.73 = x - 633
x = -44.73 + 633
x = $588.27
Approximately to the nearest dollar, the amount should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.16
is $588
