In a statistics mid-term exam graded out of 100 points, the distribution of the exam scores was bi-modal with a mean of 70 points with a standard deviation of 10 points. what percentage of students scored between 40 points and 100 points?

Step-by-step explanation:

Assuming a normal distribution for the distribution of the points scored by students in the exam, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = points scored by students

u = mean score

s = standard deviation

From the information given,

u = 70 points

s = 10.

We want to find the probability of students scored between 40 points and 100 points. It is expressed as

P(40 lesser than x lesser than or equal to 100)

For x = 40,

z = (40 - 70)/10 =-3.0

Looking at the normal distribution table, the corresponding z score is 0.0135

For x = 100,

z = (100 - 70)/10 =3.0

Looking at the normal distribution table, the corresponding z score is 0.99865

P(40 lesser than x lesser than or equal to 100) = 0.99865 - 0.0135 = 0.98515

The percentage of students scored between 40 points and 100 points will be 0.986 × 100 = 98.4%

a. 12.5 kg

step-by-step explanation:

f = ma

so

m = f/a = (2500 n)/(200 m/s²) = 12.5 kg

1 n = 1 kg·m/s²