Question

The weights of a large number of miniature poodles are approximately normally distributed with a mean of 8 kilograms and a standard deviation of 0.9 kilogram. If measurements are recorded to the nearest tenth of a kilogram, find the fraction of these poodles with weights

Answer 1

Answer and Step-by-step explanation:

Considering the table attached.

(a) over 9.5 kg;

μ = 8

σ = 0.9

z = 9.5 - 8/0.9 ≈ 1.67

P (Z > 1.67) = 0.5 - P(0<Z<1.67) = 0.5 - 0.4525 = 0.0475

(b) at most 8.6 kg;

z = 8.6-8/0.9 ≈ 0.67

P(Z < 0.67) = 0.5 + P(0<Z<0.67) = 0.5 + 0.2486 = 0.7486

(c) between 7.3 and 9.1 kg.

z₁ = 7.3-8/0.9 ≈ -0.78

z₂ = 9.1 - 8/0.9 ≈ 1.22

P(-0.78 < Z < 1.22) = P(0 < Z < 0.78) + P(0 < Z < 1.22) = 0.2823 + 0.3888 = 0.6711

Answer 2

Using the normal distribution, it is found that:

The z-score is:

$\frac{X - \mu}{0.9}$

• The probability of finding a value less than X is the p-value of Z.

• The probability of finding a value more than X is the p-value of Z.
• The probability of finding a value between a and b is the p-value of Z when X = b subtracted by the p-value of Z when X = a.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  
• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• Mean of 8, thus $\mu = 8$
• Standard deviation of 0.9, thus $\sigma = 0.9$.

The z-score will be given by:

$\frac{X - \mu}{0.9}$

Applying the percentile concept:

• The probability of finding a value less than X is the p-value of Z.

• The probability of finding a value more than X is the p-value of Z.
• The probability of finding a value between a and b is the p-value of Z when X = b subtracted by the p-value of Z when X = a.

A similar problem is given at https://brainly.com/question/13708668