Question

In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was $44.60 with a standard deviation of $2.81. The weekly amounts are approximately bell- shaped.
a. Estimate the percentage of amounts that are between $36.17 and $44.60.
b. Estimate the percentage of amounts that are between $41.79 and $47.41.
c. Estimate the percentage of amounts that are at least $38.98.
d. Between what two values will approximately 95% of the amounts be?

Answer 1

Answer:

a) the percentage of amounts that are between $36.17 and $44.60 is 49.87%

b.) Therefore  the percentage of amounts that are between $41.79 and $47.41 is 68.27%

c)  Therefore percentage of amounts that are at least $38.98 = 97.72%

d) The two values will approximately 95% of the amounts be $39.1 and $50.1

Step-by-step explanation:

i) In a recent survey of 655 working Americans, ages 25 - 34, the average weekly amount spent on lunch was $44.60 with a standard deviation of $2.81

ii) mean = $44.60

iii) standard deviation = $2.81

a) Estimate the percentage of amounts that are between $36.17 and $44.60.

   therefore we have to find P($36.27 < X < $44.60) or

P(( (36.17 - 44.60)/2.81) < z < ( (44.60 - 44.60)/2.81)) = P(-3 < Z < 0)

  therefore P(-3 < Z < 0) = P(Z < 0) - P( Z < -3) = 0.4987

 Therefore  the percentage of amounts that are between $36.17 and $44.60 is 49.87%

b)estimate the percentage of amounts that are between $41.79 and $47.41.

P($41.79 < X < $47.41) or

P(( (41.79 - 44.60)/2.81) < z < ( (47.41 - 44.60)/2.81)) = P(-1 < Z < 1)

  therefore P(-1 < Z < 1) = P(Z < 1) - P( Z < -1) = 0.6827

 Therefore  the percentage of amounts that are between $41.79 and $47.41 is 68.27%

c)Estimate the percentage of amounts that are at least $38.98.

   therefore we have to find

     P(X ≥ $38.98) or

    P(z ≥ (38.98 -  44.6)/2.81 ) = P( Z  ≥  -2)

  therefore we have to find 1 - P(Z < -2) = 1 - 0.0228 = 0.9772

 

Therefore percentage of amounts that are at least $38.98 = 97.72%

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

The two values will approximately 95% of the amounts be $39.1 and $50.1

Answer 2

Using the Empirical Rule, it is found that:

a) 49.85% of amounts are between $36.17 and $44.60.

b) 68% of amounts are between $41.79 and $47.41.

c) 97.5% of amounts are at least $38.98.

d) 95% of the amounts are between $38.98 and $50.22.

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The Empirical Rule states that:

• In a normal distribution, approximately 68% of the distribution is within 1 standard deviation of the mean.

• In a normal distribution, approximately 95% of the distribution is within 2 standard deviations of the mean.

• And that 99.7% of the measures are within 3 standard deviations of the mean.

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In this problem:

• The mean is $44.60.
• The standard deviation is $2.81.

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Question a:

• $44.60 is the mean.
• $44.6 - 3(2.81) = 36.17$, and thus, $36.17 is 3 standard deviations below the mean.
• The normal distribution is symmetric, thus, 50% of the measures are below the mean and 50% are above.
• Of those 50% below the mean, 99.7% are within 3 standard deviations of the mean, that is, between $36.17 and $44.60.

Thus:

$p = 0.5(0.997) = 0.4985$

0.4985 = 49.85%

49.85% of amounts are between $36.17 and $44.60.

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Question b:

• $41.79 is one standard deviation below the mean.
• $47.41 is one standard deviation above the mean.
• Percentage within 1 standard deviation of the mean, thus:

68% of amounts are between $41.79 and $47.41.

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Question c:

• $38.98 is two standard deviations below the mean.
• Of the 50% of amounts below the mean, 95% are greater than $38.98.
• Of the 50% of amounts above the mean, all are greater than $38.98.

Thus:

$p = 0.5(0.95) + 0.5 = 0.975$

0.975 = 97.5%

97.5% of amounts are at least $38.98.

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Question d:

• 95% of the amounts are within 2 standard deviations of the mean, thus:

$44.6 - 2(2.81) = 38.98$

$44.6 + 2(2.81) = 50.22$

95% of the amounts are between $38.98 and $50.22.

A similar problem is given at https://brainly.com/question/24244232