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Step2247 [10]
2 years ago
11

In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was 43.76 with standard deviati

on 2.67. The weekly amounts are approximately bell-shaped. (c) Between what two values will approximately 95% of the amounts be?
Mathematics
1 answer:
klemol [59]2 years ago
8 0

Answer:

Between 38.42 and 49.1.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 43.76, standard deviation of 2.67.

Between what two values will approximately 95% of the amounts be?

By the Empirical Rule, within 2 standard deviations of the mean. So

43.76 - 2*2.67 = 38.42

43.76 + 2*2.67 = 49.1

Between 38.42 and 49.1.

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I WILL AWARD BRAINLIEST IF YOU ANSWER ALL OF THESE!! THANK YOU!!
vaieri [72.5K]

Answer:

<u>1. The number is 71.3</u>

<u>2. The number is 127.1</u>

<u>3. The number is 230</u>

<u>4. The number is 34</u>

Step-by-step explanation:

Let's find out the numbers, x, for each case:

1. 5% of it is 5% of it is 23% of 15.5

5% = 0.05

23% of 15.5 = 0.23 * 15.5 = 3.565

Then, we have:

0.05x = 3.565

x = 3.565/0.05

x = 71.3

<u>The number is 71.3</u>

2. 4% of it is 31% of 16.4

4% = 0.04

31% of 16.4 = 0.31 * 16.4 = 5.084

Then, we have:

0.04x = 5.084

x = 5.084/0.04

x = 127.1

<u>The number is 127.1</u>

3. 4.5% of it is 23% of 45

4.5% = 0.045

23% of 45 = 0.23 * 45 = 10.35

Then, we have:

0.045x = 10.35

x = 10.35/0.045

x = 230

<u>The number is 230</u>

4. 5.5% of it is 17% of 11

5.5% = 0.055

17% of 11 = 0.17 * 11 = 1.87

Then, we have:

0.055x = 1.87

x = 1.87/0.055

x = 34

<u>The number is 34</u>

8 0
3 years ago
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