Question

The scores on a standardized math test for 8th grade children form a normal distribution with a mean of 80 and a standard deviation of 10. (8 points) 26. a. What proportion of the students have scores less than X- 83? b. If samples of n 6 are selected from the population, what proportion of the samples have means less than M 83? c. If samples of n 30 are selected from the population, what proportion of the samples will have means less than M83?

Answer 1

Here, we are given the distribution as:

$X \sim N(\mu = 80, \sigma = 10 )$

a) The required proportion here is computed as:

P(X < 83 )

Converting this to a standard normal variable, we get:

$P(Z < \frac{83 - 80}{10})$

Getting this from the standard normal tables, we get:

Therefore 0.6179 is the required probability here.

b) The distribution for sample means is given as:

$\bar X \sim N(\mu = 80, \sigma = \frac{10}{\sqrt{n}} )$

$\bar X \sim N(\mu = 80, \sigma = \frac{10}{\sqrt{6}} )$

$\bar X \sim N(\mu = 80, \sigma = 4.08)$

The required probability here is computed as:

$P( \bar X < 83 )$

Converting this to a standard normal variable, we get:

83-80 PIZ4.08

P(Z < 0.7353)

Getting it from the standard normal tables, we get:

P(Z < 0.7353) = 0.7689

Therefore 0.7689 is the required proportion / probability here.

c) The distribution for sample mean here is given as:

$\bar X \sim N(\mu = 80, \sigma = \frac{10}{\sqrt{30}} )$

$\bar X \sim N(\mu = 80, \sigma = 1.83)$

Therefore, the required proportion here is computed as:

$P(Z < \frac{83 - 80}{1.83})$

Getting this from the standard normal tables, we get:

Therefore 0.9498 is the required probability here.