Answer:
95% confidence interval for the difference between the average mass of eggs in small and large nest is between a lower limit of 0.81 and an upper limit of 2.39.
Step-by-step explanation:
Confidence interval is given by mean +/- margin of error (E)
Eggs from small nest
Sample size (n1) = 60
Mean = 37.2
Sample variance = 24.7
Eggs from large nest
Sample size (n2) = 159
Mean = 35.6
Sample variance = 39
Pooled variance = [(60-1)24.7 + (159-1)39] ÷ (60+159-2) = 7619.3 ÷ 217 = 35.11
Standard deviation = sqrt(pooled variance) = sqrt(35.11) = 5.93
Difference in mean = 37.2 - 35.6 = 1.6
Degree of freedom = n1+n2 - 2 = 60+159-2 = 217
Confidence level = 95%
Critical value (t) corresponding to 217 degrees of freedom and 95% confidence level is 1.97132
E = t×sd/√(n1+n2) = 1.97132×5.93/√219 = 0.79
Lower limit = mean - E = 1.6 - 0.79 = 0.81
Upper limit = mean + E = 1.6 + 0.79 = 2.39
95% confidence interval for the difference in average mass is (0.81, 2.39)
Answer is 12240 watts of power that a 12 square meter solar panel absorb.
Answer:
900
Step-by-step explanation:
.6 * 1500
Answer:
e=49
f=131
d=90
Step-by-step explanation:
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
