Question

1.Suppose that scores on a knowledge test are normally distributed with a mean of 71 and a standard deviation of 6

Answer 1

Answer:

a) For this case we can see the distribution in the figure attached is a bell shaped graph and symmetrical around 71

b) $z = \frac{76-71}{6}= 0.833$

We can see the value of 76 labeled in the second picture attached

c) $P(X>76)$

And using the z score we have this using the normal standard table or excel:

$P(Z>0.833) = 1-P(Z$

d) $n = 185*0.202= 154.16$

We can say that about 154 and 155 students scored higher than Angelica

Step-by-step explanation:

We know that X represent the random variable scores of knowledge test and is given by:

$X \sim N (\mu = 71, \sigma =6)$

Part a

For this case we can see the distribution in the figure attached is a bell shaped graph and symmetrical around 71

Part b

For this case the z score is given by:

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

And replacing we got:

$z = \frac{76-71}{6}= 0.833$

We can see the value of 76 labeled in the second picture attached

Part c

We want this probability:

$P(X>76)$

And using the z score we have this using the normal standard table or excel:

$P(Z>0.833) = 1-P(Z$

Part d

For this case we can find the number desired like this:

$n = 185*0.202= 154.16$

We can say that about 154 and 155 students scored higher than Angelica