Mathematics, 30.06.2021 15:30, davelopez979

The manager of a new supermarket wished to estimate the likely expenditure of his customers. A sample of till slips from a similar supermarket describing the weekly amount spent by 500 randomly selected customers was collected and analysed. This expenditure was found to be approximately normally distributed with a mean of $50 and a standard deviation of $15. Find the probability that any shopper selected at random spends more than $80 per week?
Find the percentage of shoppers who are expected to spend between $30 and 80 per week?

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answered: zackcarlson9374

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

88.54% of shoppers are expected to spend between $30 and 80 per week.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of $50 and a standard deviation of $15.

This means that $\mu = 50, \sigma = 15$

Find the probability that any shopper selected at random spends more than $80 per week?

This is 1 subtracted by the p-value of Z when X = 80. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{80 - 50}{15}$

Z = 2

Z = 2 has a p-value of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that any shopper selected at random spends more than $80 per week.

Find the percentage of shoppers who are expected to spend between $30 and 80 per week?

The proportion is the p-value of Z when X = 80 subtracted by the p-value of Z when X = 30.

X = 80

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{80 - 50}{15}$

Z = 2

Z = 2 has a p-value of 0.9772

X = 30

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30 - 50}{15}$

Z = -1.33

Z = -1.33 has a p-value of 0.0918

0.9772 - 0.0918 = 0.8854

0.8854*100% = 88.54%

88.54% of shoppers are expected to spend between $30 and 80 per week.

answered: Guest

if you was gonna solve for a, a is a < -12

answered: Guest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )

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The manager of a new supermarket wished to estimate the likely expenditure of his customers. A sampl...