A population of values has a normal distribution with $$ \mu=112.9 $$ and $$ \sigma=29.4 $$. You intend to draw a random sample of size $$ n=10 $$. Find the probability that a single randomly selected value is less than 142.7 . $$ P(X

Question

A population of values has a normal distribution with and . You intend to draw a
random sample of size .
Find the probability that a single randomly selected value is less than 142.7 .

Find the probability that a sample of size is randomly selected with a mean less than 142.7.

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact -scores or -
scores rounded to 3 decimal places are accepted.

Ask by Love Reid. in the United States

Mar 27,2025

UpStudy AI · Accepted Answer

We begin by noting that the population is normally distributed with mean 
$$
\mu=112.9
$$
and standard deviation 
$$
\sigma=29.4.
$$

### 1. Probability that a single value is less than 142.7

For a single value $$ X $$ from the distribution, we standardize using the $$ z $$-score:
$$
z=\frac{X-\mu}{\sigma}=\frac{142.7-112.9}{29.4}.
$$
First, compute the numerator:
$$
142.7-112.9=29.8.
$$

Then,
$$
z=\frac{29.8}{29.4}\approx1.0136.
$$

The probability that $$ X $$ is less than 142.7 is then:
$$
P(X<142.7)=P(Z<1.0136).
$$
Looking up (or using a calculator for) the standard normal cumulative distribution for $$ z=1.0136 $$ gives approximately:
$$
P(X<142.7)\approx0.8442.
$$

### 2. Probability that a sample of size $$ n=10 $$ has a mean less than 142.7

For a sample of size $$ n=10 $$, the sampling distribution of the mean $$ M $$ is normally distributed with mean 
$$
\mu_M=\mu=112.9,
$$
and standard error (SE)
$$
\sigma_M=\frac{\sigma}{\sqrt{n}}=\frac{29.4}{\sqrt{10}}.
$$

Calculate the standard error:
$$
\sigma_M = \frac{29.4}{\sqrt{10}} \approx \frac{29.4}{3.1623} \approx 9.297.
$$

Standardize the sample mean $$ M $$:
$$
z=\frac{M-\mu_M}{\sigma_M}=\frac{142.7-112.9}{9.297}.
$$
Again, the numerator is:
$$
142.7-112.9=29.8,
$$
so
$$
z=\frac{29.8}{9.297}\approx3.206.
$$

The probability that the sample mean is less than 142.7 is:
$$
P(M<142.7)=P(Z<3.206).
$$

Using the standard normal distribution, a $$ z $$-score of approximately 3.206 gives a probability very near 1, specifically:
$$
P(M<142.7)\approx0.9993.
$$

### Final Answers

$$
P(X<142.7)=0.8442 \quad \text{and} \quad P(M<142.7)=0.9993.
$$
$$ P(X<142.7)=0.8442 \quad \text{and} \quad P(M<142.7)=0.9993. $$

Solución de inteligencia artificial de Upstudy

Responder

Solución

1. Probability that a single value is less than 142.7

2. Probability that a sample of size has a mean less than 142.7

Final Answers

Beyond the Answer

preguntas relacionadas

Latest Statistics Questions