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Vadim26 [7]
3 years ago
10

A manufacturer produces a large number of microwaves. From past experience, the manufacturer knows that approximately 1.5% of al

l their microwaves are defective. Consumer Reports randomly selects 20 of these microwaves for testing. We want to determine the probability that no more than one of these microwaves is defective?
a) What is the probability that exactly one of the microwaves is defective?

b) What is the probability that at most two of the microwaves are defective?

c) Find the mean and standard deviation.
Mathematics
1 answer:
yawa3891 [41]3 years ago
4 0
This problem represents "binomial probability," because any given "experiment" can have only two possible outcomes:  defective or not defective.

Here the "population" probability that a given microwave unit is defective is 0.015.  We arbitrarily define "defective microwave" as a "success" and "non-defective microwave" as a "failure."

We can obtain binomial probability values from a calculator such as the TI-83, from a table or from the binomial probabilty formula.  In all of these cases the number of samples is given and is n=20; the probability of "success" is also given and is p=0.015.

Case 1:  What is the probability that exactly 1 microwave unit is defective?
Using the TI-83:     binompdf(20,0.015,1) = 0.225, or 9/40.  

Case 2:  What is the p. that at most 2 are defective?  Add together the binomial probabilities binompdf(20,0.015,0),binompdf(20,0.015,1),binompdf(20,0.015,2).

Result:  0.7391 + 0.2250 + 0.0326 = 0.9967.

Mean:  The mean of a binomial probability such as this one is simply np, which in his case is 20(0.015)=0.30.  Find and apply the formula for the standard deviation.
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3 years ago
The national average for the math portion of the College Board’s Scholastic Aptitude Test
Alex73 [517]

Answer:

a) 0.1587

b) 0.023

c) 0.341

d) 0.818

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 515

Standard Deviation, σ = 100

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(score greater than 615)

P(x > 615)

P( x > 615) = P( z > \displaystyle\frac{615 - 515}{100}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 615) = 1 - 0.8413 = 0.1587 = 15.87\%

b) b) P(score greater than 715)

P(x > 715) = P(z > \displaystyle\frac{715-515}{100}) = P(z > 2)\\\\P( z > 2) = 1 - P(z \leq 2)

Calculating the value from the standard normal table we have,

1 - 0.977 = 0.023 = 2.3\%\\P( x > 715) = 2.3\%

c) P(score between 415 and 515)

P(415 \leq x \leq 515) = P(\displaystyle\frac{415 - 515}{100} \leq z \leq \displaystyle\frac{515-515}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%

P(415 \leq x \leq 515) = 34.1\%

d) P(score between 315 and 615)

P(315 \leq x \leq 615) = P(\displaystyle\frac{315 - 515}{100} \leq z \leq \displaystyle\frac{615-515}{100}) = P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.841 - 0.023 = 0.818 = 81.8\%

P(315 \leq x \leq 615) = 81.8\%

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Need help on math You have a meal at a restaurant. The sale tax is 8% . You leave a tip for the waitress that is 20% of the pret
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Answer: $18

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As per given ,

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∵ Amount spent =  $23.04

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1.28x=23.04\\\\\Rightarrow\ x=\dfrac{23.04}{1.28}\\\\\Rightarrow\ x=18

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