Question

The mean amount purchased by each customer at Churchill’s Grocery Store is $33 with a standard deviation of $9. The population is positively skewed. For a sample of 56 customers, answer the following questions:

a. What is the likelihood the sample mean is at least $34? (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 $30 but less than $34? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.)

Sample mean:

c. Within what limits will 98% of the sample means occur? (Round the final answers to 2 decimal places.)

Sample mean: ___and____

Answer 1

a)

X ~ N ( µ = 33 , σ = 9 )
P ( X > 34 ) = 1 - P ( X < 34 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 34 - 33 ) / ( 9 / √ ( 56 ) )
Z = 0.83
P ( ( X - µ ) / ( σ / √ (n)) > ( 34 - 33 ) / ( 9 / √(56) )
P ( Z > 0.83 )
P ( X̅ > 34 ) = 1 - P ( Z < 0.83 )
P ( X̅ > 34 ) = 1 - 0.7967
P ( X̅ > 34 ) = 0.2033

b)

P ( 30 < X̅ < 34 ) = ?
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 30 - 33 ) / ( 9 / √(56))
Z = -2.49
Z = ( 34 - 33 ) / ( 9 / √(56))
Z = 0.83
P ( -2.49 < Z < 0.83 )
P ( 30 < X̅ < 34 ) = P ( Z < 0.83 ) - P ( Z < -2.49 )
P ( 30 < X̅ < 34 ) = 0.7967 - 0.0064
P ( 30 < X̅ < 34 ) = 0.7903

c)

P ( a < X < b ) = 0.98
Dividing the area 0.98 in two parts we get 0.98/2 = 0.49
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.49
Area above the mean is b = 0.5 + 0.49
Looking for the probability 0.01 in standard normal table to calculate Z score = -2.3263
Looking for the probability 0.99 in standard normal table to calculate Z score = 2.3263
Z = ( X - µ ) / ( σ / √(n) )
-2.3263 = ( X - 33 ) / ( 9/√(56) )
a = 30.20
2.3263 = ( X - 33 ) / ( 9/√(56) )
b = 35.80

Sample mean 30.20 and 35.80