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Assoli18 [71]
2 years ago
5

Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat

ed to loose keys (27%) or improper assembly (73%). Electrical connect defects are caused by defective wires (35%), improper connections (13%), or poorly welded wires (52%). a) Find the probability that a failure is due to loose keys. b) Find the probability that a failure is due to improperly connected or poorly welded wires.
Mathematics
2 answers:
777dan777 [17]2 years ago
8 0

Answer:

Probability that a failure is due to loose keys = 0.2376

Probability that a failure is due to improperly connected or poorly welded wires  0.078

Step-by-step explanation:

Given data:

P(A) = percentage of faulty electrical connections is 12% = 0.12

P(B) = mechanical defects is 88% =  0.88

P(C / B ) = percentage of defects that are related to loose keys is 27% =  0.27

P( D/B ) = improper assembly is 73% = 0.73

P(E/A) =  defective wires  is 35% =  0.35

P( F/A)= improper connections is 13% =  0.13

P( G/A ) = percentage of poorly welded wire is 52% = 0.52

a) Probability of being failure because of loose keys = 0.27\times 0.88 = 0.2376

b) Improperly connected = 0.13\times 0.12 = 0.0156

    poorly welded wires= 0.52 \times 0.12 = 0.0624

Probability of being failure because of improperly connected or poorly welded wires = 0.0156 + 0.0624 = 0.078

Margaret [11]2 years ago
6 0

Answer:

a) 23.76%

b) 7.8%

Step-by-step explanation:

a) probability that a failure is due to loose keys.

loose key failure (27%) comes under mechanical failure(88%)

hence, probability that a failure is due to loose keys= 0.27×0.88= 0.2376= 23.76%

b)  probability that a failure is due to improperly connected wire which comes under electrical failure = 0.12×0.13

probability that a failure is due to poorly welded wires which comes under electrical failure= 0.52×0.12

now, the probability that a failure is due to improperly connected or poorly welded wires. = 0.12(0.52+0.13)= 0.078= 7.8%

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According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating
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Answer:

a) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

H_0: \mu=32.48\\\\H_a:\mu< 32.48

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565

The degrees of freedom for this sample size are:

df=n-1=65-1=64

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

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b) 12% of peanuts are brown, 23% are blue. So 35% are either blue or brown. The probability of a peanut being blue or brown is, therefore 35/100 = 0.35.

c) 12% of peanuts are red, so the probability of a peanut being red is 12/100 = 0.12. In order to calculate the probability of 2 peanuts being both red, we can assume that the proportion doesnt change dramatically after removing one peanut (because the number of peanuts is absurdly high. We can assume that we are replenishing the peanuts). To calculate the probability of 2 peanuts being both red, we need to power 0.12 by 2, hence the probability is 0.12² = 0.0144.

d) Again, we will assume that the probability doesnt change, because we replenish. The probability of a peanut being blue is 0.23. The probability of it not being blue is 0.77, so the probability of 6 peanuts not being blue is obtained from powering 0.77 by 6, hence it is 0.77⁶ = 0.2084

e) The event 'at least one peanut is blue' is te complementary event of 'none peanuts are blue', so the probability of this event is 1- 0.2084 = 0.7916

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