Solved on Feb 11, 2024

The average grocery spending for Palestinian households is 600 NIS/week with a standard deviation of 120 NIS. What is the percentage of households spending more than 850 NIS/week? What is the probability a household spends 600 NIS/week? What is the maximum spending for a thrifty shopper household (1st quartile)?
$5.94\%$ , 1, 519.6 NIS

STEP 1

Assumptions
1. The average spending on groceries for Palestinian households is 600 NIS/week.
2. The standard deviation of spending on groceries is 120 NIS.
3. The amount of spending on groceries follows a normal distribution.
4. We need to find the percentage of households spending more than 850 NIS/week.
5. We need to find the probability of a household spending exactly 600 NIS/week.
6. We need to find the maximum amount a thrifty shopper household spends, which is in the 1st quartile of the distribution.

STEP 2

To find the percentage of households spending more than 850 NIS/week, we need to calculate the z-score for 850 NIS.
$z = \frac{X - \mu}{\sigma}$

STEP 3

Plug in the values for X (850 NIS), μ (600 NIS), and σ (120 NIS) to calculate the z-score.
$z = \frac{850 - 600}{120}$

STEP 4

Calculate the z-score.
$z = \frac{250}{120} \approx 2.0833$

STEP 5

Use the standard normal distribution table or a calculator to find the probability corresponding to a z-score of 2.0833.

STEP 6

Since we want the percentage of households spending more than 850 NIS, we need to find the area to the right of the z-score.
$P(X > 850) = 1 - P(Z \leq 2.0833)$

STEP 7

Look up the cumulative probability for $Z \leq 2.0833$ in the standard normal distribution table or use a calculator.

STEP 8

Subtract the cumulative probability from 1 to find the percentage of households spending more than 850 NIS.

STEP 9

To find the probability that a household spends exactly 600 NIS/week on groceries, we need to understand that the probability of observing a single value in a continuous distribution is zero.

STEP 10

For the maximum amount a thrifty shopper household spends (1st quartile), we need to find the z-score that corresponds to the 25th percentile (1st quartile) in the standard normal distribution.

STEP 11

Look up the z-score for the 25th percentile in the standard normal distribution table or use a calculator.

STEP 12

Once we have the z-score for the 25th percentile, we can use it to find the spending amount using the formula:
$X = \mu + z \cdot \sigma$

STEP 13

Plug in the values for μ (600 NIS), σ (120 NIS), and the z-score for the 25th percentile to calculate the maximum amount for a thrifty shopper household.

STEP 14

Calculate the maximum amount for a thrifty shopper household.
Now, let's perform the calculations:

STEP 15

Calculate the z-score for 850 NIS.
$z = \frac{850 - 600}{120} = \frac{250}{120} \approx 2.0833$

STEP 16

Using the standard normal distribution table or a calculator, find $P(Z \leq 2.0833)$ .

STEP 17

Assuming $P(Z \leq 2.0833) \approx 0.9812$ (from the standard normal distribution table or calculator), calculate the percentage of households spending more than 850 NIS.
$P(X > 850) = 1 - P(Z \leq 2.0833) = 1 - 0.9812 = 0.0188$

STEP 18

Convert the probability to a percentage.
$P(X > 850) = 0.0188 \times 100\% \approx 1.88\%$

STEP 19

Since the probability of a household spending exactly 600 NIS/week on groceries is a single point on a continuous distribution, the probability is 0.

STEP 20

Find the z-score for the 25th percentile. Assuming it is approximately -0.6745 (from the standard normal distribution table or calculator).

STEP 21

Calculate the maximum amount for a thrifty shopper household using the z-score for the 25th percentile.
$X = 600 + (-0.6745) \cdot 120$

STEP 22

Calculate the maximum amount for a thrifty shopper household.
$X = 600 - 0.6745 \cdot 120 \approx 600 - 80.94 \approx 519.06$
The percentage of households who spend more than 850 NIS/week is approximately $1.88\%$ .
The probability that a household spends exactly 600 NIS/week on groceries is 0.
The maximum amount of spending for a thrifty shopper household is approximately 519.06 NIS.

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STEP 1

STEP 2

To find the percentage of households spending more than 850 NIS/week, we need to calculate the z-score for 850 NIS.z=X−μσ z = \frac{X - \mu}{\sigma} z=σX−μ​

STEP 3

Plug in the values for X (850 NIS), μ (600 NIS), and σ (120 NIS) to calculate the z-score.z=850−600120 z = \frac{850 - 600}{120} z=120850−600​

STEP 4

Calculate the z-score.z=250120≈2.0833 z = \frac{250}{120} \approx 2.0833 z=120250​≈2.0833

STEP 5

Use the standard normal distribution table or a calculator to find the probability corresponding to a z-score of 2.0833.

STEP 6

Since we want the percentage of households spending more than 850 NIS, we need to find the area to the right of the z-score.P(X>850)=1−P(Z≤2.0833) P(X > 850) = 1 - P(Z \leq 2.0833) P(X>850)=1−P(Z≤2.0833)

STEP 7

Look up the cumulative probability for Z≤2.0833 Z \leq 2.0833 Z≤2.0833 in the standard normal distribution table or use a calculator.

STEP 8

Subtract the cumulative probability from 1 to find the percentage of households spending more than 850 NIS.

STEP 9

To find the probability that a household spends exactly 600 NIS/week on groceries, we need to understand that the probability of observing a single value in a continuous distribution is zero.

STEP 10

For the maximum amount a thrifty shopper household spends (1st quartile), we need to find the z-score that corresponds to the 25th percentile (1st quartile) in the standard normal distribution.

STEP 11

Look up the z-score for the 25th percentile in the standard normal distribution table or use a calculator.

STEP 12

Once we have the z-score for the 25th percentile, we can use it to find the spending amount using the formula:X=μ+z⋅σ X = \mu + z \cdot \sigma X=μ+z⋅σ

STEP 13

Plug in the values for μ (600 NIS), σ (120 NIS), and the z-score for the 25th percentile to calculate the maximum amount for a thrifty shopper household.

STEP 14

Calculate the maximum amount for a thrifty shopper household.Now, let's perform the calculations:

STEP 15

Calculate the z-score for 850 NIS.z=850−600120=250120≈2.0833 z = \frac{850 - 600}{120} = \frac{250}{120} \approx 2.0833 z=120850−600​=120250​≈2.0833

STEP 16

Using the standard normal distribution table or a calculator, find P(Z≤2.0833) P(Z \leq 2.0833) P(Z≤2.0833).

STEP 17

STEP 18

Convert the probability to a percentage.P(X>850)=0.0188×100%≈1.88% P(X > 850) = 0.0188 \times 100\% \approx 1.88\% P(X>850)=0.0188×100%≈1.88%

STEP 19

Since the probability of a household spending exactly 600 NIS/week on groceries is a single point on a continuous distribution, the probability is 0.

STEP 20

Find the z-score for the 25th percentile. Assuming it is approximately -0.6745 (from the standard normal distribution table or calculator).

STEP 21

Calculate the maximum amount for a thrifty shopper household using the z-score for the 25th percentile.X=600+(−0.6745)⋅120 X = 600 + (-0.6745) \cdot 120 X=600+(−0.6745)⋅120

STEP 22

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To find the percentage of households spending more than 850 NIS/week, we need to calculate the z-score for 850 NIS.
$z = \frac{X - \mu}{\sigma}$

Plug in the values for X (850 NIS), μ (600 NIS), and σ (120 NIS) to calculate the z-score.
$z = \frac{850 - 600}{120}$

Calculate the z-score.
$z = \frac{250}{120} \approx 2.0833$

Since we want the percentage of households spending more than 850 NIS, we need to find the area to the right of the z-score.
$P(X > 850) = 1 - P(Z \leq 2.0833)$

Look up the cumulative probability for $Z \leq 2.0833$ in the standard normal distribution table or use a calculator.

Once we have the z-score for the 25th percentile, we can use it to find the spending amount using the formula:
$X = \mu + z \cdot \sigma$

Calculate the maximum amount for a thrifty shopper household.
Now, let's perform the calculations:

Calculate the z-score for 850 NIS.
$z = \frac{850 - 600}{120} = \frac{250}{120} \approx 2.0833$

Using the standard normal distribution table or a calculator, find $P(Z \leq 2.0833)$ .

Convert the probability to a percentage.
$P(X > 850) = 0.0188 \times 100\% \approx 1.88\%$

Calculate the maximum amount for a thrifty shopper household using the z-score for the 25th percentile.
$X = 600 + (-0.6745) \cdot 120$