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Assume that the weights of ripe watermelons grown at a particular farm are normally distributed with a mean of 30 pounds and a standard deviation of 1.9 pounds. If the farm produces 400 ​watermelons, how many will weigh less than 27.87 ​pounds?

User Adonna
by
7.8k points

1 Answer

5 votes

Given↷

  • Mean = 30
  • Standard deviation = 1.9

  • Data value = 27.87

To find↷

  • Number of watermelons out of 400 produced melons ,which will weigh < 27.87 pounds

Answer↷

  • 52

Solution↷

We know that,


\rightarrow \: z = (x - \mu)/(\sigma) \\

  • z = standard score
  • x = measuring value
  • μ = mean
  • σ =standard derivation

inserting the given values in the formula


\rightarrow \: z = (27.87 - 30)/(1.9) \\\rightarrow \: z = - 1.121

Area that's left to z = -1.121

Area that's left to -1.121 = 0.1311

= 13.11%

Hence ,

13.11% of the watermelons weigh less than 27.87 pounds

To find the number of watermelons,

We'll find 13.11% of 400

= 13.11% x 400

= 13.11/100 x 400

= 52.44

hence , approx 52 watermelons would be lesser than 27.87 pounds

User Curie
by
7.7k points
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