33. Suppose that the scores on a statewide standardized test are normally distributed with a mean of 70 and a standard deviation of 5. Estimate the percentage of scores that were(a) between 65 and 75. %(b) above 80. %(c) below 55. %(d) between 65 and 80. %

33 Suppose that the scores on a statewide standardized test are normally distributed with a mean of 70 and a standard deviation of 5 Estimate the percentage of class=

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ANSWER:

(a) 68.27%

(b) 2.28%

(c) 0.13%

(d) 81.85%

STEP-BY-STEP EXPLANATION:

We have the following information:

[tex]\begin{gathered} m=70 \\ sd=5 \end{gathered}[/tex]

To calculate the probability we must calculate the z-value, which we do by means of the following formula:

[tex]Z=\frac{x-m}{sd}[/tex]

Then, with the value of Z and the help of the normal distribution table, we can calculate the probability.

The table is as follows:

Now, we calculate the probability in each case using the information above.

(a)

between 65 and 75:

[tex]\begin{gathered} P(65\le x\le70)=\frac{65-70}{5}\le x\le\frac{75-70}{5} \\ P(65\le x\le70)=-1\le x\le1 \\ P=0.8413-0.1587=0.6827 \end{gathered}[/tex]

68.27% between 65 and 75.

(b)

above 80:

[tex]\begin{gathered} P(x>80)=x>\frac{80-70}{5} \\ P(x>80)=x>2\rightarrow1-x<2 \\ P=1-0.9772=0.0228 \end{gathered}[/tex]

2.28% above 80.

(c)

below 55:

[tex]\begin{gathered} P(x<55)=x<\frac{55-70}{5} \\ P(x<55)=x<-3 \\ P=0.0013 \end{gathered}[/tex]

0.13% below 55.

(d)

between 65 and 80:

[tex]\begin{gathered} P(65\le x\le80)=\frac{65-70}{5}\le x\le\frac{80-70}{5} \\ P(65\le x\le80)=-1\le x\le2 \\ P=0.9772-0.1587=0.8185 \end{gathered}[/tex]

81.85% between 65 and 80.



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