Let y1,y2, . . ,yn be a random sample from a normal distribution where the mean is 2 and the variance is 4. how large must n be in order that p(1.9 ≤ x ≤ 2.1) ≥ 0.99?

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!

the answer is. -82562234434584

step-by-step explanation:

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