A group of students at Brunel university took a test in marketing and they needed to score 50 to pass. The results showed that t
he final grades have a mean of 62 and a standard deviation of 8. Suppose we approximate the distribution of these grades by normal distribution. Based on this information, you are required to perform the following; a) what percetage(to 3 decimals) of the students scored more than 78? B) what percentage of the students to 3 decimal places scored under 50? And c) what percentage to 3 decimal places of the students scored between 60 and 80?
The normal random variable of a standard normal distribution is called a standard score or a z-score. In Every normal condition, random variable X will be transformed into a z score via the following equation:
z=(X)
But when X is a normal random variable, μ is the mean of X.
For That, it is X=60
Z=(60−70)/10=−1
P(X<60)=P(Z<−1)
=0.1587
Hence percent of students failed in test is 15.87%
3.5x + 6.5y = 1800 Move 3.5x to another side. 6.5y = -3.5x +1800 Divide by 6.5. y = -3.5/6.5x + 1800 If we round it would be: Y = 0.5x + 1800 0.5x is the slope and 1800 is the y-intercept. Hope that helps.
