Let x be a random variable representing the heights of adult American men. Since it is normally distributed and the population mean and standard deviation are known, we would apply the formula,

z = (x - mean)/Standard deviation

From the information given,

mean = 68

standard deviation = 2.5

The probability that the height of a selected adult is between 63 and 73 is expressed as

$P(68\leq x\leq73)$

For x = 63,

z = (63 - 68)/2.5 = -2

Looking at the normal distribution table, the probability corresponding to the z score is 0.02275

For x = 73,

z = (73 - 68)/2.5 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.97725

Therefore,

$P(68\text{ }\leq x\leq73)\text{ = 0.97725 - 0.0}2275\text{ = 0}.9545$

Thus, the percentage of men are between 63 and 73 is

0.9545 * 100 = 95.45%

Rounding up to the nearest percentage, the answer is 95%

8 0

1 year ago

The class sizes of elementary school classes in a public school district are normally distributed with an unknown population mea

weqwewe [10]

Answer: $(17.14,\ 22.86)$

Step-by-step explanation:

The confidence interval estimate for the population mean is given by :-

$\overline{x}\pm ME$ , where $\overline{x}$ is the sample mean and ME is the margin of error.

Given : Sample mean: $\overline{x}=20$

The margin of error for a 98% confidence interval estimate for the population mean using the Student's t-distribution : $ME=2.86$

Now, the confidence interval estimate for the population mean will be :-

$20\pm2.86=(20-2.86,\ 20+2.86)=(17.14,\ 22.86)$

Hence, the 98% confidence interval estimate for the population mean using the Student's t-distribution = $(17.14,\ 22.86)$

8 0

3 years ago

Which of the following statements is true for ∠a and ∠b in the diagram?

Iteru [2.4K]

Answer:

B because they are the same angle

8 0

2 years ago

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