Answer:
1.) No ;
2.) - 0.931
3.) 0.1785
Step-by-step explanation:
Given :
μ = 84.3 ; xbar = 81.9 ; s = 17.3
H0 : μ = 84.3
H1 : μ < 84.3
The test statistic :
(xbar - μ) ÷ (s/√(n))
(81.9 - 84.3) / (17.3/√45)
-2.4 / 2.5789317
= - 0.9306
= - 0.931
Using the test statistic, we could obtain the Pvalue : df = n - 1 ; df = 45 - 1 = 44
Using the Pvalue calculator :
Pvalue(-0.9306, 44) = 0.1785
Using α = 0.05
The Pvalue > α
Then we fail to reject H0; and conclude that there is no significant evidence to support the claim that the mean waiting time is less than 84.3
A) Variance = 10.24
B) Standard Deviation = 3.2
One of the measurements of dispersion is the standard deviation, which is exclusively used for quantitative data. It aids in determining if the data's mean is a suitable measurement to reflect the core value.
TIME FREQUENCY(f) MIDPOINT(x) d d² fd fd²
0 - 0.9 43 0.45 -3 9 -129 387
1.0 - 1.9 17 1.45 -2 4 -34 68
2.0 - 2.9 19 2.45 -1 1 -19 19
3.0 - 3.9 18 3.45 0 0 0 0
4.0 - 4.9 14 4.45 1 1 14 14
5.0 -5.9 16 5.45 2 4 32 64
∑f = 127
∑fd = -136
∑fd² = 552

Standard Deviation = 

√4.35 - √1.15
Standard Deviation = 3.2
(SD)² = (3.2)² = 10.24
Variance = 10.24
To know more about standard deviation, refer to this link :
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Answer:
this is the answer unless you did it backwards
Step-by-step explanation:
3.467406380027739e-4
Answer:
378
Step-by-step explanation:
9*6=54 then 54/2=27. fnally 27*14=378
20 times 4 will get to 80