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NISA [10]
3 years ago
14

Chucky grabbed 11 1111 items in the grocery store that each had a different price and had a mean cost of about $ 4.44 $4.44dolla

r sign, 4, point, 44. On his way to the register, he gave in to an impulse to add a 1 2 th 12 th 12, start superscript, start text, t, h, end text, end superscript item: an entire wheel of cheese that cost $ 39.99 $39.99dollar sign, 39, point, 99
Mathematics
2 answers:
maria [59]3 years ago
7 0

Question:

Chucky grabbed 11 items in the grocery store that each had a different price and had a mean cost of about $4.44. On his way to the register, he gave in to an impulse to add a 12th item: an entire wheel of cheese that cost $39.99.

How will adding the wheel of cheese affect the mean and median?

Answer:

There will be a big difference in the mean when the new item is added

Step-by-step explanation:

Given

Items = 11

Mean = \$4.44

Before we solve further, we need to first calculate the total amount of the 11 items.

Mean = \frac{Total}{Items}

Make Total the Subject of formula

Total = Mean * Items

Total = \$4.44 * 11

Total = \$48.84

When the 12th item of $39.99 is added, the new mean becomes.

New\ Mean = \frac{Total + \$39.99}{11 + 1}

New\ Mean = \frac{\$48.84 + \$39.99}{11 + 1}

New\ Mean = \frac{\$88.83}{12}

New\ Mean = \$7.4025

By comparing this the old mean, we can see a huge increment between $4.44 and $7.4025

This means that there will be a big difference in the mean when the new item is added.

For the Median:

The old mean shows that the prices of the 11 items is within a small range from $4.44.

So, when the new item is added the median will only change a little bit.

In other words, the median value will only change a little bit.

ycow [4]3 years ago
4 0

Answer:

Step-by-step explanatThat is the correct anwserion:

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The first part of the question is missing and it says;

Use these parameters: Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.

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B) Percentage of men meeting the height requirement = 0.875%

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Step-by-step explanation:

From the question,

For men;

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Thus, the probability of only men will take the form of;

P(-2.36 < Z < 3.36) = P(Z<3.36) - P(Z > - 2.36)

From the normal probability table attached, when we interpolate, we'll arrive at P(Z<3.36) = 0.99961

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The original question cam be framed as;

P(62 < X < 78).

So thus, the probability of only women will take the form of;

P(-0.59 < Z < 4.93) = P(Z<4.93) - P(Z > - 0.59)

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P(Z<4.93) - P(Z > - 0.59) =0.9999996 - 0.277595 = 0.00875

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Let's find the z-value with a right-tail of 10%. From the second table i attached ;

invNorm(0.90) = 1.2816

Thus, the corresponding women's height:: x = (1.2816 x 2.9) + 63.7= 67.42 inches

For men;

We have seen that,

invNorm(0.90) = 1.2816

Thus ;

Thus, the corresponding men's height:: x = (1.2816 x 2.8) + 68.6 = 72.19 inches

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