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2 years ago

90 POINTS!! PLEASE HELP, IF YOU GIVE CORRECT ANSWER ILL GIVE YOU BRAINLIEST!!

Mathematics

1 answer:

pickupchik [31]2 years ago

7 0

Answer:

B) 2.19 %

Step-by-step explanation:

The rule of 70 is used to estimate the time (in years) it takes for an investment to double.

Rule of 70 Formula

$\textsf{Number of Years to Double}=\sf \dfrac{70}{\textsf{Annual Growth Rate}}$

Given:

Number of Years to Double = 32

Substituting this into the formula and solving

$\implies \sf 32=\sf \dfrac{70}{\textsf{Annual Growth Rate}}$

$\implies \textsf{Annual Growth Rate}=\sf \dfrac{70}{32}$

$\implies \textsf{Annual Growth Rate}=\sf 2.1875$

$\implies \textsf{Annual Growth Rate}=\sf 2.19 \% \: (nearest\:hundredth)$

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(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

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Step-by-step explanation:

The complete question is: In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below. (a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. (Round to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. (Round to four decimal places as needed.) (c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = the heights of men in the 20-29 age group

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean height = 67.5 inches

$\sigma$ = standard deviation = 2 inches

So, X ~ Normal( $\mu=67.5, \sigma^{2}=2^{2}$ )

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

P(X < 67 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{67-67.5}{2}$ ) = P(Z < -0.25) = 1 - P(Z $\leq$ 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X $\leq$ 67 inches)

P(X < 71 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{71-67.5}{2}$ ) = P(Z < 1.75) = 0.9599

P(X $\leq$ 67 inches) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{67-67.5}{2}$ ) = P(Z $\leq$ -0.25) = 1 - P(Z < 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

P(X > 71 inches) = P( $\frac{X-\mu}{\sigma}$ > $\frac{71-67.5}{2}$ ) = P(Z > 1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.9599 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

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