Question

9b )
suppose a normal distribution has a mean of 38 and a standard deviation of 2. what is the probability that a data value is between 36 and 41? round your answer to the nearest tenth of a percent.

choice

a. 83.5%
b. 62.5%
c. 77.5%
d. 56.5%
9c )
suppose a normal distribution has a mean of 38 and a standard deviation of 2. what is the probability that a data value is between 36 and 43? round your answer to the nearest tenth of a percent.

choice

a. 56.5%
b. 83.5%
c. 62.5%
d. 77.5%

Explanation:

Step1: Calcular los valores z

Para \(x = 36\): \(z_1=\frac{36 - 38}{2}=- 1\)
Para \(x = 41\): \(z_2=\frac{41 - 38}{2}=1.5\)
Para \(x = 43\): \(z_3=\frac{43 - 38}{2}=2.5\)

Step2: Usar la tabla normal estándar

Para \(z=-1\), \(P(Z < - 1)=0.1587\)
Para \(z = 1.5\), \(P(Z < 1.5)=0.9332\)
Para \(z = 2.5\), \(P(Z < 2.5)=0.9938\)

Step3: Calcular las probabilidades

Para el intervalo \(36\) a \(41\): \(P(-1<Z<1.5)=P(Z < 1.5)-P(Z < - 1)=0.9332 - 0.1587 = 0.7745\)
Para el intervalo \(36\) a \(43\): \(P(-1<Z<2.5)=P(Z < 2.5)-P(Z < - 1)=0.9938 - 0.1587 = 0.8351\)

Answer:

9b) C. \(77.5\%\)
9c) B. \(83.5\%\)