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:
okay
Step-by-step explanation:
2,200.413 is 2,165.413 to its nearest hundredth.
Answer:

Step-by-step explanation:

= 91
*
= 