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3 years ago

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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 point

s with a standard deviation of 10 points. What percentage of students scored between 40 points and 100 points?

2 answers:

MrRissso [65]3 years ago

5 0

Step-by-step explanation:

At least 89%

.............................

ur welcome

Mnenie [13.5K]3 years ago

4 0

Answer:

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%

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Step-by-step explanation:

Given

Let:

$C = Children; S = Students; A = Adults$

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For the tickets sold, we have:

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Half as many as adults are children implies that:

$A = \frac{C}{2}$

Required

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The equations to solve are:

$C + S + A = 329$ -- (1)

$5C + 7S + 12A = 2388$ -- (2)

$A = \frac{C}{2}$ -- (3)

Make C the subject in (3)

$C = 2A$

Substitute $C = 2A$ in (1) and (2)

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$2A + S + A = 329$

$3A + S = 329$

Make S the subject

$S = 329 - 3A$

$5C + 7S + 12A = 2388$ -- (2)

$5*2A + 7S + 12A = 2388$

$10A + 7S + 12A = 2388$

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Substitute $S = 329 - 3A$

$7(329 - 3A) + 22A = 2388$

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$A = 2388 - 2303$

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$C = 2 * 85$

$C = 170$

Recall that: $S = 329 - 3A$

$S = 329 - 3 * 85$

$S = 329 - 255$

$S = 74$

Hence, the result is:

$C = 170$

$S = 74$

$A = 85$

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