Question

Scores for men on the verbal portion of the SAT-I test are normally distributed with a mean of 509 and a standard deviation of 112 .
(a) If 1 man is randomly selected, find the probability that his score is at least 587 .

(b) If 20 men are randomly selected, find the probability that their mean score is at least 587 .

20 randomly selected men were given a review course before taking the SAT test. If their mean score is 587 , is there a strong evidence to support the claim that the course is actually effective?
(Enter YES or NO)

Answer 1

Answer:

a) $P(X>587)=P(Z>0.696)=1-P(Z$

b) $P(\bar X >587)=P(Z>3.11)=1-P(Z$

c) $z=\frac{587-509}{\frac{112}{\sqrt{20}}}=3.11$    

$p_v =P(Z>3.11)=0.00094$    

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean  score is significantly higher than 509 at 5% of signficance.    

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we 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. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Part a

Let X the random variable that represent the scores for the SAT-I of a population, and for this case we know the distribution for X is given by:

$X \sim N(509,112)$  

Where $\mu=509$ and $\sigma=112$

We are interested on this probability

$P(X>587)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>587)=P(\frac{X-\mu}{\sigma}>\frac{587-\mu}{\sigma})=P(Z>\frac{587-509}{112})=P(Z>0.696)$

And we can find this probability on this way:

$P(Z>0.696)=1-P(Z$

3) Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

We can find the individual probabilities like this:

$P(\bar X >587)=P(Z>\frac{587-509}{\frac{112}{\sqrt{20}}}=3.11)$

And using a calculator, excel or the normal standard table we have that:

$P(Z>3.11)=1-P(Z$

4) Part c

We need to conduct a hypothesis in order to check if  the course is actually effective:    

Null hypothesis:$\mu \leq 509$    

Alternative hypothesis:$\mu > 509$    

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$z=\frac{587-509}{\frac{112}{\sqrt{20}}}=3.11$    

P-value    

Since is a two-sided test the p value would be:    

$p_v =P(Z>3.11)=0.00094$    

Conclusion    

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean  score is significantly higher than 509 at 5% of signficance.