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
1 answer:
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 =
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Answer:
42.22% probability that the weight is between 31 and 35 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
What is the probability that the weight is between 31 and 35 pounds
This is the pvalue of Z when X = 35 subtracted by the pvalue of Z when X = 31. So
X = 35
has a pvalue of 0.5557
X = 31
has a pvalue of 0.1335
0.5557 - 0.1335 = 0.4222
42.22% probability that the weight is between 31 and 35 pounds
Answer:
x=1
y=-4
Step-by-step explanation:
- find x by making y to 0
- sub x into equation
- find y
