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MA_775_DIABLO [31]

3 years ago

11

April recieved $1,800 from her grandmother as a birthday gift. She decided to put the money in an account to save for college. I

f the account earns 9% interest compounded annually how much money will be in the account after 5 years?

1 answer:

Katarina [22]3 years ago

6 0

A = 1800(1 + (0.09 × 5)) = 2610
<span>A = $2,610.00</span>

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A normal population has a mean of 55 and a standard deviation of 14. You select a random sample of 25. Compute the probability t

Sophie [7]

Answer:

(1) The probability that the sample mean is Greater than 59 = 0.778

(2) The probability that the sample mean is less than 59 = .3632

(3) The probability that the sample mean is is Between 54 and 59 = 0.559

Step-by-step explanation:

Given -

Mean $(\nu )$ = 55

Standard deviation $(\sigma )$ = 14

Sample size ( n ) = 25

the probability that the sample mean $(\overline{X})$ is Greater than 59 =

$P(\overline{X} > 59 )$ = $P(\frac{\overline{X} - \nu}{\frac{\sigma}{\sqrt{n}}} > \frac{59 - 55 }{\frac{14}{\sqrt{25}}})$

= $P(Z > 1.428 )$

= 1 - $P(Z < 1.428 )$

= 1 - 0.9222 = .0778

the probability that the sample mean $(\overline{X})$ is Less than 54 =

$P(\overline{X}< 54 )$ = $P(\frac{\overline{X} - \nu}{\frac{\sigma}{\sqrt{n}}} < \frac{54 - 55 }{\frac{14}{\sqrt{25}}})$

= $P(Z< -.357 )$

= .3632

the probability that the sample mean is Between 54 and 59 =

$P(54< \overline{X}< 59)$ = $P(\frac{54 - 55 }{\frac{14}{\sqrt{25}}}< \frac{\overline{X} - \nu}{\frac{\sigma}{\sqrt{n}}} < \frac{59 - 55 }{\frac{14}{\sqrt{25}}})$

= $P(-.357< Z< 1.428)$

= $P(Z< 1.428 ) - P(< -.357)$

= .9222 - .3632

= 0.559

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3 years ago