Simplify 175 over 100

Mathematics

2 answers:

Ivahew [28]3 years ago

5 0

I think it would be 87 over 50 when u simplify 175 over 100

Nataly [62]3 years ago

4 0

7over4 is the simplified form

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Consider a parent population with mean 75 and a standard deviation 7. The population doesn’t appear to have extreme skewness or

Aleks04 [339]

Answer:

a) $\bar X \sim N(\mu=375, \sigma={\bar X}=\frac{7}{\sqrt{40}}=1.107)$

b) Since the sample size is large enough n>30 and the original distribution for the random variable X doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

c) $P(\bar X \leq 77)=P(Z$

d) See figure attached

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable of interest. We know from the problem that the distribution for the random variable X is given by:

$E(X) = 75$

$sd(X) = 7$

We take a sample of n=40 . That represent the sample size

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=375, \sigma={\bar X}=\frac{7}{\sqrt{40}}=1.107)$

Part b

Since the sample size is large enough n>30 and the original distribution for the random variable X doesn’t appear to have extreme skewness or outliers, the distribution for the sample mean would be bell shaped and symmetrical.

Part c

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$