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Elden [556K]
3 years ago
11

There are 42 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time

needed to grade a randomly chosen first examination paper is a random variable with an expected value of 5 min and a standard deviation of 6 min. (Give answers accurate to 3 decimal places.)
(a) If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?
1
(b) If the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?
2

Mathematics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

A) 0.99413

B) 0.00022

Step-by-step explanation:

A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:

Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes

Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:

μ = n*μ_s ample = 42 × 5 = 210 minutes

While the standard deviation for the population would be:

σ = √nσ_sample = √(42 × 6) = 15.8745 minutes

To find the z-score, we will use the formula;

z = (x - μ)/σ

Thus;

z = (250 - 210)/15.8745

z = 2.52

From the z-distribution table attached, we have;

P(Z < 2.52) ≈ 0.99413

B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.

Thus, total time is now 250 + 10 = 260 minutes

Similar to the z-formula in A above, we have;

z = (260 - 210)/15.8745

z = 3.15

P(Z > 3.15) = 0.00022

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5) Two machines M1, M2 are used to manufacture resistors with a design
Basile [38]

Answer:

Since M1 has the higher probability of being in the desired range, we choose M1.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Two machines M1, M2 are used to manufacture resistors with a design specification of 1000 ohm with 10% tolerance.

So we need the machines to be within 1000 - 0.1*1000 = 900 ohms and 1000 + 0.1*1000 = 1100 ohms.

For each machine, we need to find the probabilty of the machine being in this range. We choose the one with the higher probability.

M1:

Resistors of M1 are found to follow normal distribution with mean 1050 ohm and standard deviation of 100 ohm. This means that \mu = 1050, \sigma = 100

The probability is the pvalue of Z when X = 1100 subtracted by the pvalue of Z when X = 900. So

X = 1100

Z = \frac{X - \mu}{\sigma}

Z = \frac{1100 - 1050}{100}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

X = 900

Z = \frac{X - \mu}{\sigma}

Z = \frac{900 - 1050}{100}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

0.6915 - 0.0668 = 0.6247.

M1 has a 62.47% probability of being in the desired range.

M2:

M2 are found to follow normal distribution with mean 1000 ohm and standard deviation of 120 ohm. This means that \mu = 1000, \sigma = 120

X = 1100

Z = \frac{X - \mu}{\sigma}

Z = \frac{1100 - 1000}{120}

Z = 0.83

Z = 0.83 has a pvalue of 0.7967.

X = 900

Z = \frac{X - \mu}{\sigma}

Z = \frac{900 - 1000}{120}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033

0.7967 - 0.2033 = 0.5934

M2 has a 59.34% probability of being in the desired range.

Since M1 has the higher probability of being in the desired range, we choose M1.

8 0
3 years ago
Write five expressions that are equivalent to 20x+100
Phoenix [80]
4(5x+25)
5(4x+20)
10(2x+10)
2(10x+50)
20(x+5)

hope this helps
4 0
3 years ago
Let Y1 and Y2 have the joint probability density function given by:
Ann [662]

Answer:

a) k=6

b) P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

Step-by-step explanation:

a) if

f (y1, y2) = k(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1,  0, elsewhere

for f to be a probability density function , has to comply with the requirement that the sum of the probability of all the posible states is 1 , then

P(all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 =  ∫∫k(1 − y2) dy1*dy2 = k  ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k  ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)-  (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

then

k/6 = 1 → k=6

b)

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)=  9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

8 0
3 years ago
Ticketd to the school play cost $11 for adults and $6 for children. What expression can be used to find the toyal cost for x adu
Makovka662 [10]

Answer:221

Step-by-step explanation:

4 0
3 years ago
What is the equation for a quadratic function
GarryVolchara [31]
F(x) = ax^2 + bx + c
8 0
3 years ago
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