Another Math problem !!

A new curing process developed for a certain type of cement results in a mean compressive strength of 5000 kilograms per square

Sedbober [7]

Answer:

$\alpha =0.0668$

Step-by-step explanation:

Data given and notation

The info given by the problem is:

$n=25$ the random sample taken

$\mu =5000$ represent the population mean

$\sigma =100$ represent the population standard deviation

The critical region on this case is $\bar X$ so then if the value of $\bar X \geq 4970$ we fail to reject the null hypothesis. In other case we reject the null hypothesis

Null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the true mean is 5000, the system of hypothesis would be:

Null hypothesis: $\mu = 5000$

Alternative hypothesis: $\mu \neq 5000$

Let's define the random variable X ="The compressive strength".

We know from the Central Limit Theorem that the distribution for the sample mean is given by:

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

Find the probability of committing a type I error when H0 is true.

The definition for type of error I is reject the null hypothesis when actually is true, and is defined as $\alpha$ the significance level.

So we can define $\alpha$ like this:

$\alpha= P(Error I)= P(\bar X$

And in order to find this probability we can use the Z score given by this formula:

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

And the value for the probability of error I is givn by:

$\alpha= P(\bar X$

4 0

1 year ago

Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skat

BartSMP [9]

Let x be a random variable representing the number of skateboards produced
a.) P(x ≤ 20,555) = P(z ≤ (20,555 - 20,500)/55) = P(z ≤ 1) = 0.84134 = 84.1%

b.) P(x ≥ 20,610) = P(z ≥ (20,610 - 20,500)/55) = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%

c.) P(x ≤ 20,445) = P(z ≤ (20,445 - 20,500)/55) = P(z ≤ -1) = 1 - P(z ≤ 1) = 1 - 0.84134 = 0.15866 = 15.9%

5 0

2 years ago

A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea

uysha [10]

Answer:

The 90th percentile for the distribution of the total contributions is $6,342,525.

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 $s = \frac{\sigma}{\sqrt{n}}$ .