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 =
