The points of 100 children in a test are normally distributed with mean of 25 points and a standard deviation of 10 points.
(a)Given that the passing mark is 20, estimate the number of children who passed the test.
(b)If 5.59% of the children obtain a distinction by scoring x marks or more, estimate the value of x.
(c)If a sample of 20 children is selected, find the probability that the mean of the sample will be less than 27 points.
a) Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Multiply by 100.
b) same table, using .0559 to find Z. Insert into above equation to calculate raw score.
c) Z = (score-mean)/SEm
SEm = SD/√n
Use same table to find probability from Z.
To solve these questions, we will use the properties of the normal distribution. Let's tackle each question step by step:
(a) To estimate the number of children who passed the test, we need to calculate the probability that the score of a child is greater than or equal to the passing mark. Using the properties of the normal distribution, we can calculate this probability by standardizing the passing mark and using a standard normal distribution table.
First, we standardize the passing mark using the formula z = (x - mean) / standard deviation, where x is the passing mark, mean is the mean score, and the standard deviation is given. In this case, x = 20, mean = 25, and standard deviation = 10. Plugging in these values, we get z = (20 - 25) / 10 = -0.5.
Next, we look up the probability of a score being greater than or equal to -0.5 in the standard normal distribution table. The table gives us the cumulative probability, so we subtract the probability value from 1 to get the probability of passing.
For z = -0.5, the table gives us a cumulative probability of 0.3085. So, the probability of passing is 1 - 0.3085 = 0.6915 (approximately).
To estimate the number of children who passed the test, we multiply the probability of passing by the total number of children (100). Therefore, the estimated number of children who passed the test is 0.6915 * 100 = 69.15 (approximately).
(b) To estimate the value of x for obtaining a distinction, we need to find the score corresponding to the top 5.59% of the distribution. We can calculate this using the inverse of the cumulative distribution function (CDF) of the normal distribution.
Using the CDF table, we find the z-score corresponding to the cumulative probability of 0.9441 (1 - 0.0559). The closest z-score to this probability is approximately 1.64.
Now, we can calculate x by reversing the standardization process: x = mean + (z * standard deviation). Plugging in the values, we get x = 25 + (1.64 * 10) = 41.4 (approximately).
Therefore, the estimated value of x for obtaining a distinction is 41.4.
(c) To find the probability that the mean of a sample of 20 children will be less than 27 points, we need to use the properties of the sampling distribution of the sample mean.
The mean of the sampling distribution of sample means is equal to the population mean, which is 25 in this case. The standard deviation of the sampling distribution (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size.
Let's calculate the standard error: standard error = standard deviation / √(sample size) = 10 / √20 ≈ 2.236.
Next, we need to standardize the value of 27 using the formula z = (x - mean) / standard error. Plugging in the values, we get z = (27 - 25) / 2.236 = 0.894.
Using the standard normal distribution table, we can find the probability (area under the curve) for a z-score of 0.894. The table gives us a cumulative probability of approximately 0.8124.
Therefore, the probability that the mean of a sample of 20 children will be less than 27 points is approximately 0.8124.