Answer to Question #173861 in Statistics and Probability for Rustom

Question

In a Science test, the mean score is 42 and the standard deviation is 5. Assuming the scores are normally distributed, what is the probability that the score is:

C. between 30 and 48?

1. Convert the raw score of 48 to a z-score.

2. Draw the normal curve and locate the given z-value or values at the base line of the curve. Then, draw a vertical line through the given z-value or values and shade the required region.

3. Use the table and find the area that corresponds to the computed z-score.

4. Examine the shaded region and make an appropriate operation to apply, if needed.

5. Make a concluding statement.

Answer 1

Given, Mean score $\mu=42$

Standard deviation $\sigma=5$

(C) $P(30<X<48)=P(\dfrac{30-42}{5}<z<\dfrac{48-42}{5})=P(-2.4<z<1.2)$

$=P(0<z<1.2)+P(0<z<2.4)\\=0.3849+04918=0.8767$

1.$z=\dfrac{48-42}{5}=\dfrac{6}{5}=1.2$

2.

3.Area corrospondz to z score $P(z\leq 1.2)=0.3849$

4.No opreation is needed to apply.

5.Hence The Shaded region gives the area of 0.3849.