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USPshnik [31]
3 years ago
13

For a sample of n=64​, the probability of a sample mean being less than 20.5 if u = 21 and sigma = 1.31 is ​(Round to four decim

al places as​ needed.) Would the given sample mean be considered​ unusual?
Mathematics
1 answer:
Svetlanka [38]3 years ago
6 0

Answer:

The mean is usual

Step-by-step explanation:

We have that the mean (m) is equal to 21, the standard deviation (sd) = 1.31 and the sample size (n) = 64

They ask us for P (x <20.5)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / (sd / (n ^ 1/2))

We have all these values, replacing we have:

z = (20.5 - 21) / (1.31 / (64 ^ 1/2))

z = -3.05

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <-3.05) = 0.0002

The mean is usual because P (x> 20.5) = 1 - P (x <20.5) = 1 - 0.0002 = 0.9998 is a fairly high probability.

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2 years ago
when the wimen sells 90 oranges Rs160 with dicount of 20%,how many oranges she sell by Rs112 with profit of 20%​
Fynjy0 [20]

She sold 42 oranges with profit 20%

Step-by-step explanation:

The woman sells 90 oranges by RS.160

This price with discount of 20%

We need to find how many oranges she sell by Rs.112 with profit of 20%​

  • Selling price = cost price - discount ⇒ discount case
  • Selling price = cost price + profit ⇒ profit case

Assume that cost price of an orange is x

∵ She sells 90 oranges with Rs.160

- Find the selling price of 1 orange by dividing 160 by 90

∴ The selling price of an orange = \frac{160}{90}=\frac{16}{9}

∵ She sold them with discount 20%

∵ The cost price of an orange is x

- Find the 20% of x

∵ Discount = 20% × x = \frac{20}{100} × x = 0.2x

- To find the selling price subtract 0.2x from x

∴ The selling price = x - 0.2x

∴ The selling price of an orange = 0.8x

- Equate 0.8x by \frac{16}{9} to find x

∵ 0.8x = \frac{16}{9}

- Divide both sides by 0.8

∴ x = \frac{20}{9}

Assume that the number of oranges is y

∵ She sell y oranges for Rs.112 with profit 20%

∵ The cost price of an orange is \frac{20}{9}

- Find 20% of \frac{20}{9}

∵ Profit = 20% × \frac{20}{9} = \frac{20}{100}.\frac{20}{9}=\frac{4}{9}

- Add \frac{4}{9} to cost price to find the selling price

∴ The selling price of an orange = \frac{20}{9} + \frac{4}{9}

∴ The selling price of an orange = \frac{24}{9}

- To find y divide 112 by selling price of an orange

∵ y = 112 ÷ \frac{24}{9}

∴ y = 42

She sold 42 oranges with profit 20%

Learn more:

You can learn more about percentage in brainly.com/question/12284722

#LearnwithBrainly

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3 years ago
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