Question

Suppose that the weight of sweet cherries is normally distributed with mean μ=6 ounces and standard deviation σ=1.4 ounces. What proportion of sweet cherries weigh more than 4.7 ounces?

• Round your answer to four decimal places.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 6

standard deviation = $\sigma$ = 1.4

P(x > 4.7) = 1 - P(x < 4.7)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (4.7 - 6) / 1.4]

= 1 - P(z < -0.9286)

= 1 - 0.1765

= 0.8235

proportion = 0.8235