10x – 2 – 6x = 3x

We are given that the replacement times for the model laptop of concern are normally distributed with a mean of 3.3 years and a standard deviation of 0.6 years.

He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.1 years.

Let M = sample mean replacement time

The z-score probability distribution for sample mean is given by;

Z = $\frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ ~ N(0,1)

where, $\mu$ = population mean replacement time = 3.3 years

$\sigma$ = standard deviation = 0.6 years

n = sample of laptops = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is given by = P(M $\leq$ 3.1 years)

P(M $\leq$ 3.1 years) = P( $\frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$ $\frac{ 3.1-3.3}{\frac{0.6}{\sqrt{50} } }} }$ ) = P(Z $\leq$ -2.357) = 1 - P(Z $\leq$ 2.357)

= 1 - 0.99078 = 0.0092 or 0.92%

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.357 in the z table which will lie between x = 2.35 and x = 2.36 which has an area of 0.99078.

Hence, the required probability is 0.0092 or 0.92%.

Now, based on the result above; Yes, the computer store has been given laptops of lower than average quality because the probability of this data is unlikely to have occurred by chance alone as the probability of happening the given event is very low as 0.92%.

8 0

3 years ago

What is the total surface area of the prism? * 5.2 in 3 in 9.7 in Your answer

marissa [1.9K]

Answer:

190.28in

Step-by-step explanation:

Sides= 31.2

Front and back= 58.2

Top and bottom= 100.88

Add them together= 100.88+58.2+31.2

4 0

3 years ago

The mayor of a town has proposed a plan for the annexation of a new bridge. A political study took a sample of 1000 voters in th

garri49 [273]

Answer:

$z=\frac{0.42 -0.39}{\sqrt{\frac{0.39(1-0.39)}{1000}}}=1.945$

$p_v =P(Z>1.945)=0.0259$

If we compare the p value obtained with the significance level given $\alpha=0.02$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 2% of significance the proportion of residents who favored annexation is not significantly higher than 0.39.

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

$\hat p=0.42$ estimated proportion of residents who favored annexation

$p_o=0.39$ is the value that we want to test

$\alpha=0.02$ represent the significance level

Confidence=98% or 0.98

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 0.39:

Null hypothesis: $p\leq 0.39$

Alternative hypothesis: $p > 0.39$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.42 -0.39}{\sqrt{\frac{0.39(1-0.39)}{1000}}}=1.945$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.02$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(Z>1.945)=0.0259$

If we compare the p value obtained with the significance level given $\alpha=0.02$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 2% of significance the proportion of residents who favored annexation is not significantly higher than 0.39.

4 0

3 years ago

10x – 2 – 6x = 3x – 2 + x