Question

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?

Answer 1

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,

$\rightarrow \: z = \frac{x - \mu}{\sigma}$

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

inserting the given values in the formula

$\rightarrow \: z = \frac{27.87 - 30}{1.9} \\\rightarrow \: z = - 1.121$

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