# Determine and from the given parameters of the population and sample size.

Determine and from the given parameters of e population and sample size. mu= 86, sigma = 16, n = 64 = = Suppose a simple random sample of size n = 36 is obtained from a population with mu = 71 and sigma = 6. Describe the sampling distribution of xbar. What is P (xbar > 73)? What is P (xbar 69)? What is P (69.8 < xbar < 72.5)? Choose the correct description of the shape of the sampling distribution of xbar. The distribution is skewed left The distribution is approximately normal. The distribution is uniform. The distribution is skewed right. The shape of the distribution is unknown. Find the mean and standard deviation of the sampling distribution of x-. mu-x = sigma-x = P(x->73) = (Round to four decimal places as needed.) P (x- 69) = (Round to four decimal places as needed.) P (69.8 < xbar < 72.5) = (Round to four decimal places as needed.)

General guidance

Concepts and reason
Normal distribution: Normal distribution is a continuous distribution of data that has the bell-shaped curve. The normally distributed random variable x has mean and standard deviation.
Also, the standard normal distribution represents a normal curve with mean 0 and standard deviation 1. Thus, the parameters involved in a normal distribution are mean and standard deviation.
Standardized z-score: The standardized z-score represents the number of standard deviations the data point is away from the mean.
• If the z-score takes positive value when it is above the mean (0).
• If the z-score takes negative value when it is below the mean (0).
Sampling distribution of sample mean:
The sampling distribution of the sample mean for the given sample size n consists of the collection of the means of all possible samples of size n from the population.

Fundamentals

Let , then the standard z-score is found using the formula given below:

Where X denotes the individual raw score, denotes the population mean, and denotes the population standard deviation.
The sampling distribution of the sample mean is,
Some of the formulas for finding probability are,

Procedure for finding the z-value is listed below:
1.From the table of standard normal distribution, locate the probability value.
2.Move left until the first column is reached.
3.Move upward until the top row is reached.
4.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

Step-by-step

Step 1 of 6

From the given information we have the parameters of the population as follows:
Population mean
Population standard deviation is
Sample size is
Now, the mean of the sampling distribution is,

By using the central limit theorem, the mean of the Sampling distribution is 86.

Use to compute the standard deviation of the sampling distribution.

Step 2 of 6

The standard deviation of the sampling distribution is given by,

The mean and standard deviation of the sampling distribution is 86 and 2 respectively.

By using the central limit theorem the shape of the sampling distribution of is approximately normal with mean 86 and standard deviation is 2.

Use the central limit theorem to identify the distribution of the sampling distribution.

Step 3 of 6

(a)
According to the central limit theorem the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of population distribution shape.
The mean of the sampling distribution is,

The standard deviation of the sampling distribution is given by,

Part a
The shape of the sampling distribution of is approximately normal.

By using the central limit theorem the shape of the sampling distribution of is approximately normal.

Use to compute the required probability value.

Step 4 of 6

(b)
From the given information we have,
Population mean
Population standard deviation is
Sample size is
Compute

Part b
The probability that the sample mean is greater than 73 is 0.0228.

When the population parameter values, mean 71 and the standard deviation value 6, then the probability that the sample mean is greater than 73 is 0.0228.

Use to compute the required probability value.

Step 5 of 6

(c)
Compute

Part c
The probability that the sample mean is less than or equal to 69 is 0.0228.

When the population parameter values, mean 71 and the standard deviation value 6, then the probability that the sample mean is less than or equal to 69 is 0.0228.

Use to compute the required probability value.

Step 6 of 6

(d)
Compute

Part d
The probability that the sample mean is lies between 69.8 and 72.5 is 0.8181.

When the population parameter values, mean 71 and the standard deviation value 6, then the probability that the sample mean is lies between 69.8 and 72.5 is 0.8181.

Also Read :   The phase diagram for an organic compound is shown.

The mean and standard deviation of the sampling distribution is 86 and 2 respectively.

Part a
The shape of the sampling distribution of is approximately normal.

Part b
The probability that the sample mean is greater than 73 is 0.0228.

Part c
The probability that the sample mean is less than or equal to 69 is 0.0228.

Part d
The probability that the sample mean is lies between 69.8 and 72.5 is 0.8181.

The mean and standard deviation of the sampling distribution is 86 and 2 respectively.

Part a
The shape of the sampling distribution of is approximately normal.

Part b
The probability that the sample mean is greater than 73 is 0.0228.

Part c
The probability that the sample mean is less than or equal to 69 is 0.0228.

Part d
The probability that the sample mean is lies between 69.8 and 72.5 is 0.8181.

X-N(u,0)
z=- 11-X
*~NA
P(asz sb)=P(Z )-p(z>
P(Z 2 a)=1- P(Z )>
x = 1
98= 11
o=16
n=64
μ = μ = 86
19
= 은 | 16 64
μ. = μ = 71
= 은 ” V36
P(Z 2 a)=1- P(Z )>
IL = 11
n=36
P(x>73)
P(5 >73) = P( 3 73-72 = P(z>2) Use Excel =1- P(252) =NORMSDIST(2) = 1-0.9772 = 0.0228
P(Z )>
P(x 569)
P(BS 69) = Plot in 61 136) = P(z S-2) Use Excel =NORMSDIST(-2) = 0.0228
P(asz sb)=P(Z )-p(z>
P(69.8 x 72.5)
P(69.858 S 72.5)=P 69.8-711- 72.5-71 16//36oln61136) = P(-1.25Z 51.5) (Use Excel = P(Z
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