Question

If the mean score on a creativity test is 20, the standard deviation is 5, and the distribution is normal, the percentage of people who would obtain scores between 15 and 25 is ______

Answer 1

Answer:

68.2% of people would obtain scores between 15 and 25.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 20

Standard Deviation, σ = 5

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score between 15 and 25)

$P(15 \leq x \leq 25) = P(\displaystyle\frac{15 - 20}{5} \leq z \leq \displaystyle\frac{25-20}{5}) = P(-1 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -1)\\= 0.841 - 0.159 = 0.682 = 68.2\%$

$P(15 \leq x \leq 25) = 68.2\%$

Thus, 68.2% of people would obtain scores between 15 and 25.

Answer 2

Answer:

Step-by-step explanation:

Since the scores in the creativity test are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = scores in the creativity test.

µ = mean score

σ = standard deviation

From the information given,

µ = 20

σ = 5

We want to find the probability of people who would obtain scores between 15 and 25 is It is expressed as

P(15 ≤ x ≤ 25)

For x = 15

z = (15 - 20)/5 = - 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.1587

For x = 25

z = (25 - 20)/5 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.8413

Therefore,

P(15 ≤ x ≤ 25) = 0.8413 - 0.1587

P(15 ≤ x ≤ 25) = 0.6826

Therefore, the percentage of people who would obtain scores between 15 and 25 is

0.6826 × 100 = 68.26%