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Citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm. Find the 80th percentile for the diameters of mandarin oranges, round 4 decimal places

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ajeigbeibraheem

Answer:

6.0520

Step-by-step explanation:

Given that:

Mean = 5.85

Standard deviation = 0.24

Let N denote the 80th percentile.

P(X < N) = 0.8

[tex]P(Z < \dfrac{(N - 5.85)}{0.24}) = 0.8[/tex]

Using the Excel formula: = NORM.INV(0.8,5.85,0.24)

The 80th percentile for the diameters of mandarin oranges:

= 6.0520 (to 4 decimal place)

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