Question

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?

Answer 1

To find the value of n such that P(1.9 ≤ X ≤ 2.1) ≥ 0.99, we can use the properties of the normal distribution. By standardizing the values and using a standard normal distribution table or calculator, we find that n must be approximately 65.

Step-by-step explanation:

To find the value of n such that P(1.9 ≤ X ≤ 2.1) ≥ 0.99, we can use the properties of the normal distribution.

First, we need to standardize the values 1.9 and 2.1 using the formula Z = (X - μ) / σ, where μ is the mean (2) and σ is the standard deviation (sqrt(4) = 2).

So, Z1 = (1.9 - 2) / 2 = -0.05 and Z2 = (2.1 - 2) / 2 = 0.05. We need to find the probability that the standardized variable Z lies between -0.05 and 0.05, which is the same as finding P(-0.05 ≤ Z ≤ 0.05).

Using a standard normal distribution table or calculator, we can find that this probability is approximately 0.0392. Now, we want this probability to be greater than or equal to 0.99.

So, we set up the inequality 0.0392 ≥ 0.99 and solve for n:

0.0392 = (0.99 - 1) / sqrt(n)

Simplifying, we have 0.0392 = -0.01 / sqrt(n). Cross-multiplying, we get sqrt(n) = -0.01 / 0.0392.

Taking the square root of both sides, we find that n = (-0.01 / 0.0392)^2. Evaluating this expression, we get n ≈ 64.81. Since n must be a whole number, we can round up to n = 65.

Answer 2

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] +
`Z_(1-\alpha /2)`*(σ/√n) - (Y[bar] +
`Z_(1-\alpha /2)`*(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=
`Z_(1-\alpha /2)`*(σ/√n)

d*
`Z_(1-\alpha /2)`= 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!