Who runs the fastest ?? Help

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

HELP ME PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEE

ANTONII [103]

Multiply 400 times 0.70 and that’s how you get the % (in this case 280)

5 0

3 years ago

125 tickets were sold for the jazz band concert for a total of $1,022. Students tickets cost $6 each, and general admission tick

rodikova [14]

Answer:

Number of student tickets sold are 57 and general tickets sold are 68.

Step-by-step explanation:

Let the number of student tickets = x and the number of general tickets = y.

It is given that in total 125 tickets were sold.

So, we have,

$x+y=125$

Also, students tickets is sold for $6 each and the general tickets are sold for $10 each.

Since, the total cost of the tickets is $1022.

So, we get,

$6x+10y=1022$

Thus, the system of equations obtained is,

x + y = 125 ........................(1)

6x + 10y = 1022

Multiply equation (1) by 6 gives us,

6x + 6y = 750 ............................(2)

6x + 10y = 1022 .........................(3)

Subtact (2) from (3), we get,

10y - 6y = 1022 - 750

i.e. 4y = 272

i.e. y= 68.

So, equation (1) gives,

x = 125 - y

i.e. y = 125 - 68

i.e. y = 57

Hence, the number of student tickets sold are 57 and general tickets sold are 68.

3 0

3 years ago

Which expressions are equivalent?

alexdok [17]

Answer:

I believe the one that is selected is correct.

Step-by-step explanation:

1/5k - 2/3j and -2/3j + 1/5k

If you switch the subtraction to addition, 2/3 becomes negative, while 1/5 stays the same.

4 0

2 years ago

A box contains 5 red chips, 3 blue chips, and 7 white chips. A chip is randomly selected, replaced, and a second chip is selecte

Basile [38]

So add the number of chips together
5+3+7=15
So the chance of a red chip is 5/15
Blue 3/15
White 7/15
So the blue chips would be 3/15 chance
So take away one chip 14
So 7/14
The awnsers is
The probability of selecting a blue chip will be 3/15 and for a white chip it is 7/14

5 0

3 years ago