Question

22c )
suppose a normal distribution has a mean of 34 and a standard deviation of 2. what is the probability that a data value is between 29 and 35? round your answer to the nearest tenth of a percent.

choice

a. 67.5%
b. 66.5%
c. 68.5%
d. 69.5%

Explanation:

Step1: Calcular z-scores

Para \( x = 29 \): \( z=\frac{29 - 34}{2}=\frac{- 5}{2}=-2.5 \)
Para \( x = 35 \): \( z=\frac{35 - 34}{2}=\frac{1}{2}=0.5 \)

Step2: Encontrar probabilidades de z-scores

Usando la tabla Z o calculadora:

• \( P(Z < - 2.5)\approx0.0062 \)
• \( P(Z < 0.5)\approx0.6915 \)

Step3: Calcular la probabilidad entre z-scores

\( P(-2.5 < Z < 0.5)=P(Z < 0.5)-P(Z < - 2.5)=0.6915 - 0.0062 = 0.6853\approx68.5\% \)

Answer:

C. 68.5%