Answer:
The heaviest 5% of fruits weigh more than 747.81 grams.
Step-by-step explanation:
We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.
Let X = <u><em>weights of the fruits</em></u>
The z-score probability distribution for the normal distribution is given by;
Z =
~ N(0,1)
where,
= population mean weight = 733 grams
= standard deviation = 9 grams
Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;
P(X > x) = 0.05 {where x is the required weight}
P(
>
) = 0.05
P(Z >
) = 0.05
In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;



x = 747.81 grams
Hence, the heaviest 5% of fruits weigh more than 747.81 grams.
Answer 420.
a, is the first term. d is difference.
Since this is an arithmetic progression, I use this formula.
Un = nth term
Sn = sum of terms.
First find the number of terms, which is 20.
Then use the second formula to find the sum.
There is NO MODE for the data set because none of the numbers are repeated at least once.
The answer is going to be 7,434,000. hope that helped
Answer:
Step-by-step explanation:
it would be to subract 0.01 from 0.025 because you want to get the x by itself