Question

what is the probability that a data value in a normal distribution is between a z - score of - 1.52 and a z - score of - 0.34? round your answer to the nearest tenth of a percent.
a. 27.3%
b. 29.3%
c. 30.3%
d. 28.3%
question 3
3a )
suppose a normal distribution has a mean of 26 and a standard deviation of 4. what is the probability that a data value is between 28 and 31? round your answer to the nearest tenth of a percent.
a. 19.3%
b. 21.3%
c. 20.3%
d. 22.3%

Explanation:

2c)

Step1: Encontrar la probabilidad para \(z = - 1.52\)

Usando la tabla \(z\), \(P(z<-1.52)=0.0643\)

Step2: Encontrar la probabilidad para \(z=-0.347\approx - 0.35\)

Usando la tabla \(z\), \(P(z < - 0.35)=0.3632\)

Step3: Calcular la probabilidad entre los dos \(z -\)scores

\(P(-1.52<z<-0.35)=P(z < - 0.35)-P(z < - 1.52)\)
\(P(-1.52<z<-0.35)=0.3632 - 0.0643=0.2989\)

Step4: Convertir a porcentaje

\(0.2989\times100 = 29.89\%\approx29.3\%\)

Step1: Calcular los \(z -\)scores

Para \(x = 28\): \(z_1=\frac{28 - 26}{4}=\frac{2}{4}=0.5\)
Para \(x = 31\): \(z_2=\frac{31 - 26}{4}=\frac{5}{4}=1.25\)

Step2: Encontrar las probabilidades para los \(z -\)scores

Usando la tabla \(z\), \(P(z<0.5)=0.6915\) y \(P(z < 1.25)=0.8944\)

Step3: Calcular la probabilidad entre los dos \(z -\)scores

\(P(0.5<z<1.25)=P(z < 1.25)-P(z < 0.5)\)
\(P(0.5<z<1.25)=0.8944 - 0.6915=0.2029\)

Step4: Convertir a porcentaje

\(0.2029\times100=20.29\%\approx20.3\%\)

Answer:

B. \(29.3\%\)

3a)