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

Answer

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 X-N(u,0), then the standard z-score is found using the formula given below:

z=-
11-X

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*~NA,

Some of the formulas for finding probability are,

P(asz sb)=P(Z <b)-P(Z sa)

P(Z 2 a)=1- P(Z <a)

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 98= 11

Population standard deviation is o=16

Sample size is n=64

Now, the mean of the sampling distribution is,

μ = μ = 86

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

Step 2 of 6

The standard deviation of the sampling distribution is given by,

= 은
| 16
64

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.

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,

μ. = μ = 71

The standard deviation of the sampling distribution is given by,

= 은

V36

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.

Step 4 of 6

(b)

From the given information we have,

Population mean IL = 11

Population standard deviation is

Sample size is n=36

Compute 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

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.

Step 5 of 6

(c)

Compute P(x 569)

P(BS 69) = Plot in 61 136)
= P(z S-2)
Use Excel
=NORMSDIST(-2)
= 0.0228

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.

Step 6 of 6

(d)

Compute 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 <1.5) - P(Z S-1.2) =NORMSDIST(1.5)
(=N

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.

Answer

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.

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