Answer:
W= 5.744
Step-by-step explanation:
given that a grocery store produce manager is told by a wholesaler that the apples in a large shipment have a mean weight of 6 ounces and a standard deviation of 1.4 ounces
Sample size n= 49
Margin of error = 0.10 (10% risk )
Let us assume X no of apples having mean weight of 6 oz is N(6,1.4)
Then sample mean will be normal with (6, 1.4/7) = (6,0.2)
(Because sample mean follows normal with std error as std dev /sqrt of sample size)
Now required probability <0.10
i.e.
Since x bar is normal we find z score for

From std normal distribution table we find that z = 1.28
Corresponding X score =

Answer:
A. 17.5
Step-by-step explanation:
4x+8 27
L-------------M-------------------------N
<------------------ 6x -------------------->
LM = 4x+8
MN = 27
LN = 6x
LM + MN = LN
4x+8 + 27 = 6x
35 = 2x
17.5 = x
Answer:
0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand
Step-by-step explanation:
For each customer, there are only two possible outcomes. Either they choose the leading brand, or they do not. The probability of a customer choosing the leading brand is independent of any other customer, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
The leading brand of dishwasher detergent has a 30% market share.
This means that 
A sample of 25 dishwasher detergent customers was taken.
This means that 
a. What is the probability that 10 or fewer customers choose the leading brand?
This is:

In which












Then

0.9021 = 90.21% probability that 10 or fewer customers choose the leading brand
Answer:
B
Step-by-step explanation:
Tbh I would say ounces or grams. I wouldn't make a final decision on my answer though.
._.