What involves examining and evaluating information, as well as weighing alternatives and considering consequences?
1 answer:
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Answer:
n= 60
Step-by-step explanation:
Hello!
You have Y₁, Y₂, ..., Yₙ random sample with a normal distribution: Y~N(μ;σ²)
μ= 2
σ²= 4
You need to calculate a sample size n so that (1.9 ≤ Y ≤2.1)= 0.99
To reach the sample size you need to work with the distribution of the sample mean (Y[bar]) because it is this distribution that is directly affected by the sample size.
Y[bar]~N(μ;σ²/n)
Under the sample mean distribution you have to use the standard normal:
Z= Y[bar] - μ ~N(0;1)
σ/√n
Now the asked interval is:
P(1.9 ≤ Y[bar] ≤2.1)= 0.99
The upper bond is 2.1
The lower bond is 1.9
The difference between the two bonds is the amplitude of the interval a=2.1-1.9= 0.2
And the probability included between these two bonds is 0.99
With this in mind you can rewite it as an interval for the sample mean:
Y[bar] + *(σ/√n) - (Y[bar] +
*(σ/√n))= 0.2
Using the semiamplitude (d) of the interval you can easly calculate the required sample:
d= a/2= 0.2/2= 0.1
d= *(σ/√n)
d* = σ/√n
√n*(d* [tex]Z_{1-\alpha /2}[/tex)= σ
√n= σ/(d* [tex]Z_{1-\alpha /2}[/tex)
n= (σ/(d* [tex]Z_{1-\alpha /2}[/tex))²
n= (2/(0.1* 2.586))²
n= 59,81 ≅ 60
I hope it helps!
Answer: Five cats ate 1 and a quarter of cat food.
Step-by-step explanation
Given:-
Cat food ate by one cat = 1/4 cup
Cat food ate by five cats ==1 and 1/4 cup.
Therefore cat food ate by five cats = 1 and a quarter of cup.[ here a quarter =1/4]