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!

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.