. It is known that the glucose level in blood of diabetic persons follows a normal distribution model with mean 106 mg/100 ml and standard deviation 8 mg/100 ml. a. Calculate the probability of a random diabetic person having a glucose level less than 120 mg/100 ml. b. What percentage of persons have a glucose level between 90 and 120 mg/100 ml?

Answers

Answer 1

Answer:

a. 0.9599 = 95.99% probability of a random diabetic person having a glucose level less than 120 mg/100 ml.

b. 0.9371 = 93.71% of people have a glucose level between 90 and 120 mg/100 ml.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 p-value, we get the probability that the value of the measure is greater than X.

Mean 106 mg/100 ml and standard deviation 8 mg/100 ml

This means that [tex]\mu = 106, \sigma = 8[/tex]

a. Calculate the probability of a random diabetic person having a glucose level less than 120 mg/100 ml.

This is the p-value of Z when X = 120. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{120 - 106}{8}[/tex]

[tex]Z = 1.75[/tex]

[tex]Z = 1.75[/tex] has a p-value of 0.9599

0.9599 = 95.99% probability of a random diabetic person having a glucose level less than 120 mg/100 ml.

b. What percentage of persons have a glucose level between 90 and 120 mg/100 ml?

The proportion is the p-value of Z when X = 120 subtracted by the p-value of Z when X = 90. So

X = 120

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{120 - 106}{8}[/tex]

[tex]Z = 1.75[/tex]

[tex]Z = 1.75[/tex] has a p-value of 0.9599

X = 90

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{120 - 106}{8}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a p-value of 0.0228

0.9599 - 0.0228 = 0.9371

0.9371 = 93.71% of people have a glucose level between 90 and 120 mg/100 ml.


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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answers

Answer:

The second one (answer of 3), but the other ones could've worked, they were just calculated wrong.

Step-by-step explanation:

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Answers

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Answers

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

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Answers

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Step-by-step explanation:

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Answers

All questions are using SOH from SOH CAH TOA. So the equations you’re using is Sin(degree°) = x/number given on side
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Answer:

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Answers

12 1/2 years

Step-by-step explanation:

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Answers

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Answers

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• 3) There are two choices of burritos at a restaurant, vegetarian or beef. The random variable represents the total number out of 254 customers who ordered beef.

• 4) Based on the parents' genetics, each of 6 children from a particular pair of parents has a 0.30 probability of having blue eyes. The random variable represents the total number of children from this pair of parents with blue eyes

Step-by-step explanation:

The binomial distribution simply means the probability of success or failure in an experiment. The instances in which the variable described is binomial are given below:

• 2) A quality check on a particular product must meet five guidelines. All products are made in the same factory under the same conditions. The random variable represents the total number of products out of 35 tested that pass inspection.

• 3) There are two choices of burritos at a restaurant, vegetarian or beef. The random variable represents the total number out of 254 customers who ordered beef.

• 4) Based on the parents' genetics, each of 6 children from a particular pair of parents has a 0.30 probability of having blue eyes. The random variable represents the total number of children from this pair of parents with blue eyes.

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A self-serve frozen yogurt shop has 8 candy toppings and 4 fruit toppings to choose from. How many ways are there to top a frozen yogurt?

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Answers

4096 ways : 8 candy toppings plus 4 fruit toppings is 12. Then 2^12 (2 different types of toppings and the 12 of total toppings) = 4096

Annapolis Company purchased a $4,000, 6%, 5-year bond at 101 and held it to maturity. The straight line method of amortization is used for both premiums & discounts. What is the net cash received over the life of the bond investment? (all money received minus all money paid, round to nearest whole dollar)

Answers

Answer:

The answer is "[tex]\bold{\$1160}[/tex]"

Step-by-step explanation:

Calculating total paid money:

[tex]= \$4000 \times 101\% \\\\= \$4000 \times \frac{101}{100} \\\\=\$40 \times 101\\\\=\$4040[/tex]

[tex]\text{Total received money = Principle on Maturity + Interest for 5 years}[/tex]

                                   [tex]= \$4000 + \$4000\times 6\% \times 5 \\\\= \$4000 + \$4000\times \frac{6}{100} \times 5 \\\\= \$4000 + \$40 \times 6 \times 5 \\\\= \$4000 + \$40 \times 30 \\\\= \$4000 + \$1200 \\\\= \$5200 \\\\[/tex]

Total earnings over the life of the corporate bond

[tex]= \$5200 - \$4040 \\\\=\$1160[/tex]

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