Use technology or a z-score table to answer the question.

The number of baby carrots in a bag is normally distributed with a mean of 94 carrots and a standard deviation of 8.2 carrots.

Approximately what percent of the bags of baby carrots have between 90 and 100 carrots?

46%

Step-by-step explanation:

Since the number of baby carrots in a bag is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = the number of baby carrots in the bag.

µ = mean

σ = standard deviation

From the information given,

µ = 94 carrots

σ = 8.2 carrots

The probability that a bag of baby carrots have between 90 and 100 carrots is expressed as

P(90 ≤ x ≤ 100)

For x = 90,

z = (90 - 94)/8.2 = - 0.49

Looking at the normal distribution table, the probability corresponding to the z score is 0.31

For x = 100,

z = (100 - 94)/8.2 = 0.73

Looking at the normal distribution table, the probability corresponding to the z score is 0.77

Therefore,

P(90 ≤ x ≤ 100) = 0.77 - 0.31 = 0.46

The percent of the bags of baby carrots that have between 90 and 100 carrots is

0.46 × 100 = 46%

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58