To solve this problem, we will use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Step 1: Identify the parameters.
The population mean (μ) is 90, the population standard deviation (σ) is 10, and the sample size (n) is 100.

Step 2: Calculate the standard deviation of the sample mean.
The standard deviation of the sample mean (σx̄) can be found using the formula σ/√n. Substituting our values in, we get 10/√100 = 1.

Step 3: Convert the sample mean range to a z-score.
The z-score is calculated by (X - μ) / σx̄. For X = 83, the z-score is (83 - 90) / 1 = -7. For X = 94, the z-score is (94 - 90) / 1 = 4.

Step 4: Find the probability that the z-score is between -7 and 4.
This is done by looking up these values in the z-table, which gives the probability that a value chosen at random from the distribution is less than or equal to z. The value for z = 4 is 0.99997 and for z = -7 is practically 0. Therefore, the probability that the sample mean is between 83 and 94 is 0.99997 - 0 = 0.99997 or 99.997%.

Question

To solve this problem, we will use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Step 1: Identify the parameters.
The population mean (μ) is 90, the population standard deviation (σ) is 10, and the sample size (n) is 100.

Step 2: Calculate the standard deviation of the sample mean.
The standard deviation of the sample mean (σx̄) can be found using the formula σ/√n. Substituting our values in, we get 10/√100 = 1.

Step 3: Convert the sample mean range to a z-score.
The z-score is calculated by (X - μ) / σx̄. For X = 83, the z-score is (83 - 90) / 1 = -7. For X = 94, the z-score is (94 - 90) / 1 = 4.

Step 4: Find the probability that the z-score is between -7 and 4.
This is done by looking up these values in the z-table, which gives the probability that a value chosen at random from the distribution is less than or equal to z. The value for z = 4 is 0.99997 and for z = -7 is practically 0. Therefore, the probability that the sample mean is between 83 and 94 is 0.99997 - 0 = 0.99997 or 99.997%.

Knowee AI · Accepted Answer

To solve this problem, we will use the Central Limit Theorem which states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

Step 1: Identify the parameters.
The population mean (μ) is 90, the population standard deviation (σ) is 10, and the sample size (n) is 100.

Step 2: Calculate the standard deviation of the sample mean.
The standard deviation of the sample mean (σx̄) can be found using the formula σ/√n. Substituting our values in, we get 10/√100 = 1.

Step 3: Convert the sample mean range to a z-score.
The z-score is calculated by (X - μ) / σx̄. For X = 83, the z-score is (83 - 90) / 1 = -7. For X = 94, the z-score is (94 - 90) / 1 = 4.

Step 4: Find the probability that the z-score is between -7 and 4.
This is done by looking up these values in the z-table, which gives the probability that a value chosen at random from the distribution is less than or equal to z. The value for z = 4 is 0.99997 and for z = -7 is practically 0. Therefore, the probability that the sample mean is between 83 and 94 is 0.99997 - 0 = 0.99997 or 99.997%.

An unknown distribution has a mean of 90 and a standard deviation of 10. Samples of size n = 100 are drawn randomly from the population. Find the probability that the sample mean is between 83 and 94. (3 marks)

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