21.12.2022

Find the​ z-scores for which 15​% of the​ distribution's area lies between -z and z

. 5

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Mathematics
Step-by-step answer
P Answered by PhD

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x = r) = nCr × p^r × q^(n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - p) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 36.5% = 36.5/100 = 0.365

q = 1 - p = 1 - 0.365

q = 0.635

n = 15

a) P(x = 0) = 15C0 × 0.365^0 × 0.635^(15 - 0) = 0.0011

P(x = 1) = 15C1 × 0.365^1 × 0.635^(15 - 1) = 0.0095

P(x = 2) = 15C2 × 0.365^2 × 0.635^(15 - 2) = 0.038

P(x = 3) = 15C3 × 0.365^3 × 0.635^(15 - 3) = 0.095

P(x = 4) = 15C4 × 0.365^4 × 0.635^(15 - 4) = 0.16

P(x = 5) = 15C5 × 0.365^5 × 0.635^(15 - 5) = 0.21

k P(X = k)

0 0.0011

1 0.0095

2 0.038

3 0.095

4 0.16

5 0.21

b) mean = np = 15 × 0.365 = 5.475

c) standard deviation = √npq

= √15 × 0.365 × 0.635

= 1.86

d) z = (x - mean)/standard deviation

x = 2

z = (2 - 5.475)/1.86 = - 1.87

Mathematics
Step-by-step answer
P Answered by Specialist

a) 0.1587 b) 0.023 c) 0.1587 d) 1.15 e)-0.95      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 510

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 610)

P(x > 610)

P( x  610) = P( z  \displaystyle\frac{610 - 510}{100}) = P(z  1)

= 1 - P(z \leq 1)

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

P(x  610) = 1 - 0.8413 = 0.1587 = 15.87\%

b) P(score greater than 710)

P(x  710) = P(z  \displaystyle\frac{710-510}{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  710) = 2.3\%

c)P(score between 410 and 510)

P(410 \leq x \leq 510) = P(\displaystyle\frac{410 - 510}{100} \leq z \leq \displaystyle\frac{510-510}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%

P(410 \leq x \leq 510) = 34.1\%

d) x = 625

z_{score} = \displaystyle\frac{625 - 510}{100} = \displaystyle\frac{115}{100} = 1.15

e) x = 415

z_{score} = \displaystyle\frac{415 - 510}{100} = \displaystyle\frac{-95}{100} = -0.95

Mathematics
Step-by-step answer
P Answered by PhD

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x = r) = nCr × p^r × q^(n - r)

Where

x represent the number of successes.

p represents the probability of success.

q = (1 - p) represents the probability of failure.

n represents the number of trials or sample.

From the information given,

p = 36.5% = 36.5/100 = 0.365

q = 1 - p = 1 - 0.365

q = 0.635

n = 15

a) P(x = 0) = 15C0 × 0.365^0 × 0.635^(15 - 0) = 0.0011

P(x = 1) = 15C1 × 0.365^1 × 0.635^(15 - 1) = 0.0095

P(x = 2) = 15C2 × 0.365^2 × 0.635^(15 - 2) = 0.038

P(x = 3) = 15C3 × 0.365^3 × 0.635^(15 - 3) = 0.095

P(x = 4) = 15C4 × 0.365^4 × 0.635^(15 - 4) = 0.16

P(x = 5) = 15C5 × 0.365^5 × 0.635^(15 - 5) = 0.21

k P(X = k)

0 0.0011

1 0.0095

2 0.038

3 0.095

4 0.16

5 0.21

b) mean = np = 15 × 0.365 = 5.475

c) standard deviation = √npq

= √15 × 0.365 × 0.635

= 1.86

d) z = (x - mean)/standard deviation

x = 2

z = (2 - 5.475)/1.86 = - 1.87

Mathematics
Step-by-step answer
P Answered by Master

a) 0.1587 b) 0.023 c) 0.1587 d) 1.15 e)-0.95      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 510

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 610)

P(x > 610)

P( x  610) = P( z  \displaystyle\frac{610 - 510}{100}) = P(z  1)

= 1 - P(z \leq 1)

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

P(x  610) = 1 - 0.8413 = 0.1587 = 15.87\%

b) P(score greater than 710)

P(x  710) = P(z  \displaystyle\frac{710-510}{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  710) = 2.3\%

c)P(score between 410 and 510)

P(410 \leq x \leq 510) = P(\displaystyle\frac{410 - 510}{100} \leq z \leq \displaystyle\frac{510-510}{100}) = P(-1 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1)\\= 0.500 - 0.159 = 0.341 = 34.1\%

P(410 \leq x \leq 510) = 34.1\%

d) x = 625

z_{score} = \displaystyle\frac{625 - 510}{100} = \displaystyle\frac{115}{100} = 1.15

e) x = 415

z_{score} = \displaystyle\frac{415 - 510}{100} = \displaystyle\frac{-95}{100} = -0.95

Mathematics
Step-by-step answer
P Answered by PhD

2.7 and 99.6%

Step-by-step explanation:

In this question, we are asked to calculate the percentage of the population scores as a genius.

We proceed as follows;

z =( Mean - value)/standard deviation= (140 - 100)/15 = 2.6667 = 2.7

From normal probability table, given that z = 2.67, the probability is 0.9962

This means that the required percentage is 99.6%

Mathematics
Step-by-step answer
P Answered by PhD

2.7 and 99.6%

Step-by-step explanation:

In this question, we are asked to calculate the percentage of the population scores as a genius.

We proceed as follows;

z =( Mean - value)/standard deviation= (140 - 100)/15 = 2.6667 = 2.7

From normal probability table, given that z = 2.67, the probability is 0.9962

This means that the required percentage is 99.6%

Mathematics
Step-by-step answer
P Answered by PhD
Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram
Mathematics
Step-by-step answer
P Answered by PhD

For 1 flavor there are 9 topping

Therefore, for 5 different flavors there will be 5*9 choices

No of choices= 5*9

=45 

Mathematics
Step-by-step answer
P Answered by PhD

For every 8 cars there are 7 trucks

Therefore,

Cars:Truck=8:7

Answer is B)8:7

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