5

The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 70 inches and a standard devia

Sav [38]

Answer:

The probabilty would be 87.34%

Step-by-step explanation:

Add all the numbers together and multiply them as a percetnage by four

6 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let $p_1$ be the probability that a bearing meets the specification.

So, $p_1=0.9$

Sample size, $n_1=500$ , is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: $\mu_1=n_1p_1=500\times0.9=450$

Variance: $\sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45$

$\Rightarrow \sigma_1 =\sqrt{45}=6.71$

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, $X\geq 440.$

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56$ .

So, the probability that a given shipment is acceptable is

$P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062$

Hence, the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by $p_2$ .

$p_2=0.94$

The total number of shipment, i.e sample size, $n_2= 300$

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, $\mu_2$ , and variance, $\sigma_2^2$ .

Mean: $\mu_2=n_2p_2=300\times0.94=282$

Variance: $\sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92$

$\Rightarrow \sigma_2=\sqrt(16.92}=4.11$ .

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

$z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85$ .

So, the probability that a given shipment is acceptable is

$P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977$

Hence, the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, $\alpha$ be the required probability of acceptance of one shipment.

So,

$-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}$

On solving

$\alpha= 0.977896$

Again, the probability of acceptance of one shipment, $\alpha$ , depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let $p$ be the probability that one bearing meets the specification. So

$-2.01=\frac{439.5-500 p}{\sqrt{500 p(1-p)}}$

On solving

$p=0.9053$

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0

3 years ago

Anyone mind helping me solve this inequality? Step by step

dlinn [17]

By decomposing the figure in simpler shapes, we will see that the total area is:

a = 180 cm²

<h3>How to find the area of the composite figure?</h3>

Remember that the area of a rectangle of width W and length L is:

A = L*W

And the area of a triangle with base B and height H is:

A = B*H/2.

Then, the upper part can be seen as a rectangle of length of 6cm and width of 6 cm, with two triangles on the sides, such that each triangle has a base of 3cm and a height of 6cm.

So the area of that part is:

A = 6cm*6cm + 2*(3cm*6cm/2) = 54cm²

Now, the bottom triangle has a base of 12 cm, and a height of:

15cm - 6cm = 9cm

Then its area is:

A' = 12cm*9cm/2 = 54cm²

This means that the total area of the figure is:

total area = 54cm² + 54cm² = 108cm²

If you want to learn more about area:

brainly.com/question/24487155

#SPJ1

5 0

1 year ago

Shoutout to my sister amckeon88

kotykmax [81]

Periodt Pooh. For brainly??

4 0

3 years ago

What is the sum of all the whole numbers from 1 to 1000?

Katen [24]

Answer:

So there are 500 pairs of numbers that have a sum of 1001. Thus, the sum of numbers from 1 to 1000 is 500*1001 = 500,500.

Step-by-step explanation:

5 0

3 years ago