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Raleigh went to a gaming arcade for his birthday. His parents gave him $55 to spend. He purchased $14.25 in food and drinks, and

xenn [34]

Answer:

Kindly check explanation

Step-by-step explanation:

Given that :

Total amount to spend = $55

Amount of food and drinks purchased = $14.25

Amount put on gaming card = $(55 - 14.25) = $40.75

Cost per game = $1.25

Number of games (g) he can play

Number of games g:

Amount put on gaming card / cost per game

= $40.75 / $1.25

= 32.6 games

g ≤ 32

6 0

2 years ago

Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the

Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.6, \sigma = 0.5$ .