Question

Suppose that the scores on a statewide standardized test are normally distributed with a mean of 66 and astandard deviation of 4. Estimate the percentage of scores that were(a) between 58 and 74.%(b) above 74.%(c) below 62.%(d) between 62 and 74.%

Answer 1

ANSWER:

(a) 95.44%

(b) 2.28%

(c) 15.87%

(d) 81.85%

STEP-BY-STEP EXPLANATION:

Given:

μ = 66

σ = 4

We must calculate the value of z in each case depending on the values given, we calculate the value of z as follows:

$z=\frac{x-\mu}{\sigma}$

(a) between 58 and 74.

P (58 < x < 74)

$\begin{gathered} P(58We determine these values with the table, just like this:

Therefore:

[tex]\begin{gathered} P(z<2)-P(z<-2)=0.9772-0.0228 \\ \\ P(58

(b) above 74

P (x > 74)

[tex]\begin{gathered} P(x>74)=1-P(x<74) \\ \\ P(x\gt74)=1-P\left(z<\frac{74-66}{4}\right) \\ \\ P(x\gt74)=1-P\left(z<2\right) \\ \\ P(x\gt74)=1-0.9772 \\ \\ P(x\gt74)=0.0228=2.28\% \end{gathered}$

(c) below 62.

P (x < 62)

$\begin{gathered} P(x<62)=P\left(z<\frac{62-66}{4}\right) \\ \\ P(x<62)=P\left(z<-1\right) \end{gathered}$

We determine this value with the table, like this:

Therefore:

$P(x<62)=0.1587=15.87\%$

(d) between 62 and 74.

P (62 < x < 74)

[tex]\begin{gathered} P(58