Question

use for questions 7 - 10: the weights of dogs in a dog show is normally distributed with a mean of 58 pounds and a standard deviation of 17.2 pounds. use a standard normal distribution curve to find each probability.

1. p(x > 51)

z = \frac{51 - 58}{17.2}=-0.41
p(z < - 0.41)=0.7469
=74.69(100)=66541

1. p(35 < x < 62)
2. p(x < 102)
3. p(x > 85)

Explanation:

Step1: Standardize the values

We use the formula $z=\frac{x - \mu}{\sigma}$, where $\mu = 58$ is the mean and $\sigma=17.2$ is the standard - deviation.

Step2: Calculate probabilities for each part

For $P(X > 51)$:

First, calculate the z - score: $z=\frac{51 - 58}{17.2}=\frac{-7}{17.2}\approx - 0.41$. Then, $P(X > 51)=P(Z>-0.41)$. Since $P(Z > - 0.41)=1 - P(Z\leq - 0.41)$, and from the standard normal table $P(Z\leq - 0.41)=0.3409$, so $P(X > 51)=1 - 0.3409 = 0.6591$.

For $P(35

Calculate the z - scores. For $x = 35$, $z_1=\frac{35 - 58}{17.2}=\frac{-23}{17.2}\approx - 1.34$. For $x = 62$, $z_2=\frac{62 - 58}{17.2}=\frac{4}{17.2}\approx0.23$. Then $P(35 < X < 62)=P(-1.34

For $P(X < 102)$:

Calculate the z - score: $z=\frac{102 - 58}{17.2}=\frac{44}{17.2}\approx2.56$. From the standard normal table, $P(Z < 2.56)=0.9948$.

For $P(X > 85)$:

Calculate the z - score: $z=\frac{85 - 58}{17.2}=\frac{27}{17.2}\approx1.57$. Then $P(X > 85)=P(Z > 1.57)=1 - P(Z\leq1.57)$. From the standard normal table, $P(Z\leq1.57)=0.9418$, so $P(X > 85)=1 - 0.9418 = 0.0582$.

Answer:

1. $0.6591$
2. $0.5009$
3. $0.9948$
4. $0.0582$