Question

1)

a) A manufacturer knows that their items have a normally distributed length, with a mean of 13.4 inches, and standard deviation of 2 inches.

If one item is chosen at random, what is the probability that it is less than 7.5 inches long?

b) A manufacturer knows that their items lifespans are normally distributed with mean = 14.2 and standard deviation = 3.9.

What proportion of the items' lifespans will be longer than 25 years?

c) A particular fruit's weights are normally distributed, with a mean of 538 grams and a standard deviation of 30 grams.

If you pick one fruit at random, what is the probability that it will weigh between 481 grams and 574 grams

Answer 1

1) Solution :

Given that ,

a) mean = $\mu$ = 13.4

standard deviation = $\sigma$ = 2

P(x < 7.5) = P[(x - $\mu$ ) / $\sigma$ < (7.5 - 13.4) / 2]

= P(z < -2.95)

= 0.0016

Probability = 0.0016

b) mean = $\mu$ = 14.2

standard deviation = $\sigma$ = 3.9

P(x > 25) = 1 - P(x < 25)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (25 - 14.2) /3.9 )

= 1 - P(z < 2.77)

= 1 - 0.9972

0.0028

Proportion = 0.0028

c) mean = $\mu$ = 538

standard deviation = $\sigma$ = 30

P(481 < x < 574) = P[(481 - 538)/ 30) < (x - $\mu$ ) /$\sigma$  < (574 - 538) /30 ) ]

= P(-1.9 < z < 1.2)

= P(z < 1.2) - P(z < -1.9)

= 0.8849 - 0.0287

0.8562

Probability = 0.8562