Question

The final scores of students in a graduate course are distributed normally with a mean of 72 and a standard deviation of 5. What is the probability that a student's score will be between 65 and 78?

Answer 1

Answer:

0.8041

Step-by-step explanation:

We know that  

μ=72  and  σ=5

and P(65<X<78)

We can determine the Z value as (X-μ)/σ

P( 65<X<78 )=P( 65-72< X-μ<78-72)

$P((65-72)/5 <Z<(78-72)/5)$

$P(65<X<78 )=P(-1.4<Z<1.2)$

To fine the Z values:

$P (-1.4<Z<1.2 )=P ( Z<1.2 )-P (Z<-1.4 )$

From the standard normal tables:

$P (Z<1.2 )=0.8849$

to find P ( Z<-1.4)

$P ( Z<-a)=1-P ( Z<a )$

From the standard normal tables:

$P ( Z<-1.4)=1-P ( Z<1.4 )=1-0.9192=0.0808$

Therefore

$P(-1.4<Z<1.2 )=0.8041$

$P (65<X<78)=0.8041$

Answer 2

Using the Zscore principle, the probability that a student's score will be between 65 and 78 is 0.80

Recall :

• Zscore = (score - mean) ÷ σ

For a score of 65 :

Zscore = (65 - 72) ÷ 5

Zscore = -1.4

Using the normal distribution table :

P(Z < - 1.4) = 0.08

For a score of 65 :

Zscore = (78 - 72) ÷ 5

Zscore = 1.2

Using the normal distribution table :

P(Z < 1.2) = 0.88

P(Z < 1.2) - P(Z < -1.4) = 0.88 - 0.08 = 0.8

Hence, the probability that a student's score will be between 65 and 78 is 0.80

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