Question

A grocery? store's receipts show that Sunday customer purchases have a skewed distribution with a mean of $31 and a standard deviation of $21. Suppose the store had 292 customers this Sunday.

?a) Estimate the probability that the? store's revenues were at least $9,500.

?b) If, on a typical? Sunday, the store serves 292 customers, how much does the store take in on the worst 10?% of such? days?

Answer #1

(a) = Population mean = 31

= Population SD = 21

n = sample size = 292

= sample mean = 9500/292 = 32.5342

SE = /

= 21/ = 1.2147

To find P(32.5342):

Z = ( - )/SE

= (32.5342 - 31)/1.2147 = 1.2630

Table of Area Under Standard Normal Curve gives area = 0.3962

So,

P(>32.5342) = 0.5 - 0.3962 = 0.1038

So,

Answer is:

0.1038

(b)

Worst 10% corresponds to area = 0.5 - 0.10 = 0.40 from mid value tro Z on LHS.

Table gives Z score - 1.28

So,

Z = - 1.28 = ( - 31)/1.2147

So,

= 31 - (1.28 X 1.2147)

= 31 - 1.5548

= 29.45

So,

Answer is:

29.45 X 292 = $ 8,597.99

A grocery store's receipts show that Sunday customer purchases
have a skewed distribution with a mean of $31 and a standard
deviation of $21. Suppose the store had 318 customers this Sunday.
a) Estimate the probability that the store's revenues were at
least $10 comma 600. b) If, on a typical Sunday, the store
serves 318 customers, how much does the store take in on the worst
1% of such days?

A grocery store's receipts show that Sunday customer purchases
have a skewed distribution with a mean of $31 and a standard
deviation of $22. Suppose the store had 297 customers this
Sunday.
a) Estimate the probability that the store's revenues were at
least $9 comma 800.
b) If, on a typical Sunday, the store serves 297 customers,
how much does the store take in on the worst 10% of such
days?

A grocery store's receipts show that Sunday customer purchases
have a skewed distribution with a mean of $29 and a standard
deviation of $20. Suppose the store had 293 customers this
Sunday.
a) Estimate the probability that the store's revenues were at
least $9,100.
b) If, on a typical Sunday, the store serves 293 customers,
how much does the store take in on the worst 10% of such
days?

A grocery store's receipts show that Sunday customer purchases
have a skewed distribution with a mean of $30 and a standard
deviation of $18. Suppose the store had 316 customers this Sunday.
a) Estimate the probability that the store's revenues were at
least $10 comma 100. b) If, on a typical Sunday, the store
serves 316 customers, how much does the store take in on the worst
1% of such days? Also if you know which program to use on...

A grocery store's receipts show that Sunday customer purchases
have a skewed distribution with a mean of $32 and a standard
deviation of $24. Suppose the store had 305 customers this Sunday.
a) Estimate the probability that the store's revenues were at
least $10 comma 000. b) If, on a typical Sunday, the store
serves 305 customers, how much does the store take in on the worst
10% of such days? a) The probability is... (Round to four
decimal places...

A
grocery store’s reciepts show that Sunday customer purchases bave a
skewed distribution with a mean of $34 and a standard deviation of
$22. Suppose the store had 292 customers this Sunday.
a.) Estimate the probability that the store’s revenues were at
least $10,200 (round to four decimal places as needed)
b.) The store takes in at most $_____
(round to two decimal places as needed)
for
b.)
If on a typical sunday, the store serves 292 customers, how
much...

1. Hoping to lure more shoppers downtown, a city builds a new
public parking garage in the central business district. The city
plans to pay for the structure through parking fees. For a random
sample of 42 weekdays, daily fees collected averaged $133, with
standard deviation of $16.
- Find a 95% confidence interval for the mean daily income
this parking garage will generate.
The 95% confidence interval for the mean daily income is
2. A grocery store's receipts show...

A
grocery store’s reciepts show that Sunday customer purchases have a
skewed distribution with a mean of $32 and a standard deviation of
$20. Complete parts a through c below.
-Multiple choice-
1.) Explain why you cannot determime the probability that the
next Sunday customer will spend at least $40.
a.) The probability can only be determimed if the point is
less than one standard deviation away from the mean
b.) The probability can only be determimed if the point...

Grocery store receipts show that customer purchases have a
normaldistribution with a mean of $32 and standard deviation of
$20. a) If someone’s spends one standard deviation below the mean,
how much did they spend? b) Find the 90thpercentile of purchases.c)
Find the probability that a random customer spends more than $35.
D).Find the probability that the next 50 customers will spend at
least $35 on average.

The mean amount purchased by each customer at Churchill’s
Grocery Store is $29 with a standard deviation of $8. The
population is positively skewed. For a sample of 58 customers,
answer the following questions: a. What is the likelihood the
sample mean is at least $31? (Round the z-value to 2 decimal places
and the final answer to 4 decimal places.) Sample mean b. What is
the likelihood the sample mean is greater than $26 but less than
$31? (Round...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 21 minutes ago

asked 31 minutes ago

asked 34 minutes ago

asked 43 minutes ago

asked 54 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago