Pls help someone pls pls help I have exam tomorrow

Find the surface of the triangular prism

maxonik [38]

To find the area of the triangular faces, use the formula A = 1/2bh, where A = area, b = base, and h = height

7 0

3 years ago

a population has a mean of 25 median of 24 and mode of 26 the standard deviation is 5 the value for the

Arada [10]

Complete Question

A population has a mean of 25, a median of 24, and a mode of 26. The standard deviation is 5. The value of the 16th percentile is _______. The range for the middle 3 standard deviation is _______.

Answer:

The value of the 16th percentile is $x = 20$

The range for the middle 3 standard deviation is $10 \to 40$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 25$

The median is $m = 24$

The mode is $n = 26$

The standard deviation is $\sigma = 5$

Generally the 16th percentile is mathematically represented as

$P(x)= P(\frac{X - \mu }{ \sigma } \le \frac{x- 25 }{5} ) = 0.16[/teGenerally [tex]\frac{X - \mu}{\sigma } = Z(The \ standardized \ value \ of \ X )$

So

$P(x) = P(Z \le \frac{ x - 25}{5} ) = 0.16$

Now from the normal distribution table the z-score of 0.16 is

z = -1

$P(x) = P(Z \le \frac{ x - 25}{5} ) = P(Z \le -1 )$

=> $\frac{x- 25}{5} = -1$

=> $x- 25 = -5$

=> $x = 25-5$

=> $x = 20$

The range for the middle 3 standard deviation is mathematically represented

$\mu - 3\sigma \to \mu + 3\sigma$

$25 - 3(5 ) \to 25 + 3(5 )$

$10 \to 40$

3 0

3 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$